Kinematics and Linear Motion Practice Questions — JAMB UTME

Question 1Question →

A particle starts from rest and accelerates uniformly at a rate of $4\text{ m/s}^2$ for a duration $t_1$ . Immediately after reaching its maximum velocity, it decelerates uniformly at $2\text{ m/s}^2$ until coming to rest. If the total distance covered during the entire motion is $600\text{ m}$ , what is the total time of motion in seconds?

Show answer & explanation

Answer: 30

Answer

The total time of motion is 30 seconds.

For a two-stage motion starting and ending at rest, the peak velocity is

v_{\text{max}} = a_1 t_1 = a_2 t_2

, giving a time ratio

t_2 / t_1 = a_1 / a_2 = 4 / 2 = 2

. The total distance

S

is the area under the velocity-time triangle,

S = \frac{1}{2} v_{\text{max}} (t_1 + t_2) = \frac{1}{2} (4 t_1) (3 t_1) = 6 t_1^2

. Setting

6 t_1^2 = 600

yields

t_1 = 10\text{ s}

, which gives a total time

T = t_1 + t_2 = 30\text{ s}

.

Step-by-Step Solution

1

Relate maximum velocity to the acceleration time

t_1

v_{\text{max}} = 4 t_1

Using

v = u + a t

starting from rest (

u = 0

).

2

Relate deceleration time

t_2

to

t_1

t_2 = 2 t_1

The final velocity is

0

, so

0 = v_{\text{max}} - a_2 t_2 \implies t_2 = \frac{4 t_1}{2} = 2 t_1

.

3

Express the total displacement

S

as a function of

t_1

S = 6 t_1^2

Displacement during acceleration

s_1 = \frac{1}{2}(4)t_1^2 = 2 t_1^2

. Displacement during deceleration

s_2 = \frac{1}{2}(2)(2 t_1)^2 = 4 t_1^2

. Total

S = 2 t_1^2 + 4 t_1^2 = 6 t_1^2

.

4

Solve for the acceleration time

t_1

t_1 = 10\text{ s}

Given

S = 600\text{ m}

, we have

6 t_1^2 = 600 \implies t_1^2 = 100 \implies t_1 = 10\text{ s}

.

5

Calculate the total time of motion

T

T = 30\text{ s}

Total time is the sum of both phases:

T = t_1 + t_2 = t_1 + 2 t_1 = 3 t_1 = 3(10) = 30\text{ s}

.

Key Concept

Multi-stage uniform motion and average velocity relations

Estimated Time:3m 0s

Question 2Question →

A car traveling along a straight road at an initial speed of $20\text{ m/s}$ applies its brakes, causing a uniform deceleration of $5\text{ m/s}^2$ until it comes to a complete stop. What is the total distance traveled by the car during this braking period?

Show answer & explanation

Answer: 40

Answer

The total distance traveled by the car while coming to a stop is

40\text{ m}

.

Using the third equation of linear motion

v^2 = u^2 + 2as

with

u = 20\text{ m/s}

,

v = 0\text{ m/s}

, and acceleration

a = -5\text{ m/s}^2

, we obtain

0 = 400 - 10s

, which simplifies directly to

s = 40\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic parameters from the problem statement.

u = 20\text{ m/s}

,

v = 0\text{ m/s}

,

a = -5\text{ m/s}^2

The car decelerates to a stop, so final velocity is zero and acceleration is negative relative to initial direction of motion.

2

Select the appropriate equation of motion linking

u

,

v

,

a

, and displacement

s

.

v^2 = u^2 + 2as

This equation directly relates initial velocity, final velocity, acceleration, and distance without requiring time.

3

Substitute the known values into the equation and solve for distance

s

.

0^2 = 20^2 + 2(-5)s \implies 10s = 400 \implies s = 40\text{ m}

Algebraic simplification yields the stopping distance.

Key Concept

Uniformly Accelerated Motion and Stopping Distance

Question 3Question →

A car starts from rest and accelerates uniformly to a velocity of $16\text{ m/s}$ in $4\text{ s}$ . It continues at this constant velocity for $8\text{ s}$ , and then decelerates uniformly to rest in another $4\text{ s}$ . What is the average speed of the car for the entire journey?

12\text{ m/s}

16\text{ m/s}

8\text{ m/s}

24\text{ m/s}

Show answer & explanation

Answer: $12\text{ m/s}$

Answer

12\text{ m/s}

The average speed of an object undergoing multi-stage motion is defined as the total distance traveled divided by the total time taken. The total distance covered is

32\text{ m}

(during acceleration)

+ 128\text{ m}

(during constant velocity)

+ 32\text{ m}

(during deceleration)

= 192\text{ m}

. Dividing

192\text{ m}

by the total time of

16\text{ s}

yields

12\text{ m/s}

.

Step-by-Step Solution

1

Calculate the distance covered in each stage of motion.

Acceleration stage (

s_1

):

\frac{0 + 16}{2} \times 4 = 32\text{ m}

. Constant velocity stage (

s_2

):

16 \times 8 = 128\text{ m}

. Deceleration stage (

s_3

):

\frac{16 + 0}{2} \times 4 = 32\text{ m}

.

The total distance is the sum of distances traveled during acceleration, constant speed, and deceleration.

2

Find total distance traveled and total time taken.

Total distance (

S

) =

32\text{ m} + 128\text{ m} + 32\text{ m} = 192\text{ m}

. Total time (

T

) =

4\text{ s} + 8\text{ s} + 4\text{ s} = 16\text{ s}

.

Average speed requires total distance divided by total elapsed time.

3

Compute the average speed.

Average speed =

\frac{S}{T} = \frac{192\text{ m}}{16\text{ s}} = 12\text{ m/s}

.

Dividing total distance by total time gives the average speed over the entire motion.

Key Concept

Average Speed in Multi-Stage Motion

Estimated Time:1m 30s

Question 4Question →

A ball $P$ is dropped from rest from the top of a cliff of height $100\text{ m}$ . At the same instant, another ball $Q$ is projected vertically upwards from the base of the cliff along the same vertical line with an initial speed of $50\text{ m/s}$ . Taking $g = 10\text{ m/s}^2$ , at what height above the ground do the two balls meet?

20\text{ m}

50\text{ m}

80\text{ m}

100\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

The two balls meet at a height of

80\text{ m}

above the ground.

Equating the position equations of both objects gives

100 - 5t^2 = 50t - 5t^2

, which simplifies to

50t = 100

, so

t = 2\text{ s}

. Substituting

t = 2\text{ s}

into the vertical height formula gives

h = 80\text{ m}

above the ground.

Step-by-Step Solution

1

Set up equations of motion for both balls in terms of time

t

.

For ball

P

(dropped from top): height above ground

h_P = 100 - \frac{1}{2}gt^2 = 100 - 5t^2

.
For ball

Q

(projected from ground): height above ground

h_Q = ut - \frac{1}{2}gt^2 = 50t - 5t^2

.

Both balls move simultaneously under gravity along the same vertical line.

2

Equate the heights

h_P = h_Q

to find the time of meeting

t

.

100 - 5t^2 = 50t - 5t^2 \implies 50t = 100 \implies t = 2\text{ s}

.

The balls pass each other when their heights above the ground are equal.

3

Calculate the height above ground using

t = 2\text{ s}

.

h = 50(2) - 5(2)^2 = 100 - 20 = 80\text{ m}

.

Substituting

t = 2\text{ s}

into either height expression yields the position where they meet.

Key Concept

Relative vertical motion under gravity

Alternative Method

Using relative velocity: The relative acceleration between the two balls is

g - g = 0\text{ m/s}^2

. The relative speed of approach is constant at

50\text{ m/s}

. The initial separation is

100\text{ m}

, so time to meet is

t = \frac{100}{50} = 2\text{ s}

. Height above ground is then

h = 100 - \frac{1}{2}(10)(2)^2 = 80\text{ m}

.

Estimated Time:1m 30s

Question 5Question →

A body is projected vertically upwards from the ground with an initial velocity $u$ . It passes a point at a height of $40\text{ m}$ above the ground at $t = 2\text{ s}$ while ascending and again at $t = 4\text{ s}$ while descending. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the initial speed of projection $u$ of the body?

20\text{ m/s}

30\text{ m/s}

40\text{ m/s}

60\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

The initial speed of projection of the body is

30\text{ m/s}

.

The position of a body thrown vertically upwards is given by

h = ut - \frac{1}{2}gt^2

. Rearranging this equation into standard quadratic form gives

t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

. The roots

t_1

and

t_2

correspond to the times the body reaches height

h

. By Vieta's formulas, the sum of the times is

t_1 + t_2 = \frac{2u}{g}

. Substituting

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

gives

6 = \frac{2u}{10}

, which solves to

u = 30\text{ m/s}

. Alternatively, the time to reach maximum height is the midpoint

t_{\text{top}} = \frac{t_1 + t_2}{2} = 3\text{ s}

, and at maximum height

v = 0 = u - gt_{\text{top}}

, yielding

u = 10 \times 3 = 30\text{ m/s}

.

Step-by-Step Solution

1

Set up the vertical motion displacement equation

h = ut - \frac{1}{2}gt^2

The equation describes the vertical position

h

at any time

t

for a projectile launched from ground level with initial speed

u

.

2

Rearrange the equation into standard quadratic form for

t

\frac{1}{2}gt^2 - ut + h = 0 \implies t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

The two solutions

t_1

and

t_2

represent the times at which the body reaches the specific height

h

.

3

Apply Vieta's formulas for the sum of roots of the quadratic equation

t_1 + t_2 = \frac{2u}{g}

The sum of the roots of a quadratic equation

t^2 - Bt + C = 0

is equal to the coefficient

B

.

4

Substitute given values to solve for

u

2 + 4 = \frac{2u}{10} \implies 6 = \frac{u}{5} \implies u = 30\text{ m/s}

Given

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

, direct substitution yields the initial velocity.

Key Concept

Vertical Motion under Gravity and Time Symmetry

Estimated Time:1m 0s

Question 6Question →

A body accelerates uniformly from rest at a rate of $4\text{ m/s}^2$ for $6\text{ s}$ . What is the distance covered by the body during this time interval?

72\text{ m}

144\text{ m}

24\text{ m}

12\text{ m}

Show answer & explanation

Answer: $72\text{ m}$

Answer

The distance covered by the body is

72\text{ m}

.

Applying the equation of motion

s = ut + \frac{1}{2}at^2

with

u = 0\text{ m/s}

,

a = 4\text{ m/s}^2

, and

t = 6\text{ s}

yields

s = 0 + \frac{1}{2}(4)(6^2) = 72\text{ m}

.

Step-by-Step Solution

1

Identify the given kinematic parameters

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 4\text{ m/s}^2

, and time interval

t = 6\text{ s}

.

Since the body starts from rest, its initial velocity is zero.

2

Select and set up the equation of motion for displacement

s = ut + \frac{1}{2}at^2 = (0)(6) + \frac{1}{2}(4)(6)^2

This formula directly relates displacement to initial velocity, acceleration, and time under constant acceleration.

3

Calculate the total distance

s = \frac{1}{2} \times 4 \times 36 = 72\text{ m}

Squaring

6\text{ s}

yields

36\text{ s}^2

, and multiplying by

2\text{ m/s}^2

gives

72\text{ m}

.

Key Concept

Linear motion under uniform acceleration

Estimated Time:45s

Question 7Question →

A body is released from rest from the top of a cliff of height $h$ . If it covers a distance equal to $\frac{7}{16}h$ in the final second of its motion before hitting the ground, what is the total height $h$ of the cliff? (Take acceleration due to gravity $g = 10\text{ m s}^{-2}$ )

45\text{ m}

80\text{ m}

100\text{ m}

160\text{ m}

Show answer & explanation

Answer: $80\text{ m}$

Answer

80\text{ m}

Using the equation of motion under constant acceleration, the total distance fallen from rest in time

t

is

h = \frac{1}{2}gt^2

. The distance fallen up to time

(t-1)

is

h_{t-1} = \frac{1}{2}g(t-1)^2

. The distance traveled during the final second is

\Delta h = h - h_{t-1} = \frac{1}{2}g(2t-1)

. Equating this to

\frac{7}{16}h

yields

\frac{1}{2}g(2t-1) = \frac{7}{16}\left(\frac{1}{2}gt^2\right)

, which simplifies to

7t^2 - 32t + 16 = 0

. Factoring gives

t = 4\text{ s}

(rejecting

t = \frac{4}{7}\text{ s}

since time must exceed

1\text{ s}

). Substituting

t = 4\text{ s}

into

h = \frac{1}{2}(10)(4)^2

gives

80\text{ m}

.

Step-by-Step Solution

1

Express total height

h

in terms of total fall time

t

.

h = \frac{1}{2} g t^2 = 5t^2

Since the body starts from rest (

u = 0\text{ m s}^{-1}

), displacement under uniform acceleration

g = 10\text{ m s}^{-2}

is given by

h = \frac{1}{2}gt^2

.

2

Express the height fallen in the first

(t - 1)

seconds.

h' = \frac{1}{2} g (t - 1)^2 = 5(t - 1)^2

The distance covered up to one second before impact is the total distance fallen minus the distance covered in the final second.

3

Calculate the distance fallen in the final second and set up the equation.

\Delta h = h - h' = 5t^2 - 5(t - 1)^2 = 5(2t - 1)

. Given

\Delta h = \frac{7}{16}h

, we have

5(2t - 1) = \frac{7}{16}(5t^2)

.

The distance fallen during the last second is the difference between total height and height fallen up to

(t-1)

seconds.

4

Solve the quadratic equation for

t

.

7t^2 - 32t + 16 = 0 \implies (7t - 4)(t - 4) = 0 \implies t = 4\text{ s}

(since

t > 1\text{ s}

).

Simplifying

2t - 1 = \frac{7}{16}t^2

gives

7t^2 - 32t + 16 = 0

. The root

t = 4/7\text{ s}

is discarded as

t

must be greater than

1\text{ s}

.

5

Substitute

t = 4\text{ s}

back into the total height formula.

h = 5(4)^2 = 80\text{ m}

.

Calculating total height using

h = 5t^2

for

t = 4\text{ s}

gives

80\text{ m}

.

Key Concept

Free Fall under Gravity and Motion in the

n

-th Second

Question 8Question →

A train accelerates uniformly along a straight track from an initial velocity of $10\text{ m/s}$ to a final velocity of $30\text{ m/s}$ over a distance of $100\text{ m}$ . What is the magnitude of the acceleration of the train?

0.2\text{ m/s}^2

4\text{ m/s}^2

5\text{ m/s}^2

8\text{ m/s}^2

Show answer & explanation

Answer: $4\text{ m/s}^2$

Answer

4\text{ m/s}^2

Using the equation of motion

v^2 = u^2 + 2as

, substitute

u = 10\text{ m/s}

,

v = 30\text{ m/s}

, and

s = 100\text{ m}

. This gives

30^2 = 10^2 + 2(a)(100)

, which simplifies to

900 - 100 = 200a

, or

800 = 200a

. Solving for acceleration yields

a = 4\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the given kinematic values from the problem statement.

Initial velocity

u = 10\text{ m/s}

, final velocity

v = 30\text{ m/s}

, and displacement

s = 100\text{ m}

.

Listing known values helps in selecting the appropriate equation of motion.

2

Select the linear motion formula relating

u

,

v

,

s

, and acceleration

a

.

v^2 = u^2 + 2as

This formula connects initial velocity, final velocity, distance, and acceleration without requiring time

t

.

3

Substitute the values into the equation and solve for

a

.

30^2 = 10^2 + 2(a)(100) \implies 900 = 100 + 200a \implies 800 = 200a \implies a = 4\text{ m/s}^2

Algebraic rearrangement yields the magnitude of acceleration.

Key Concept

Equations of Uniformly Accelerated Motion

Estimated Time:45s

Question 9Question →

A motorist traveling along a straight highway at a constant speed of $30\text{ m/s}$ observes a road hazard ahead. The driver experiences a reaction delay of $0.5\text{ s}$ before applying the brakes, after which the vehicle decelerates uniformly at a rate of $5\text{ m/s}^2$ . What is the total distance traveled by the vehicle from the instant the hazard is observed until the vehicle comes to a complete stop?

90\text{ m}

97.5\text{ m}

105\text{ m}

195\text{ m}

Show answer & explanation

Answer: $105\text{ m}$

Answer

The total distance traveled by the vehicle before coming to a stop is

105\text{ m}

.

The motion consists of two distinct stages: a constant-speed phase during the

0.5\text{ s}

reaction time (

15\text{ m}

) and a uniformly decelerating phase until rest (

90\text{ m}

). Summing both components yields

105\text{ m}

.

Step-by-Step Solution

1

Calculate the reaction distance (

s_1

) covered during the driver's reaction delay.

s_1 = u \times t_r = 30\text{ m/s} \times 0.5\text{ s} = 15\text{ m}

Before the brakes are applied, the vehicle continues moving at its initial constant speed of

30\text{ m/s}

.

2

Calculate the braking distance (

s_2

) using the third equation of motion.

v^2 = u^2 + 2as_2 \implies 0^2 = 30^2 + 2(-5)s_2 \implies 10s_2 = 900 \implies s_2 = 90\text{ m}

The vehicle decelerates from

30\text{ m/s}

to a final velocity of

0\text{ m/s}

at a uniform deceleration rate of

a = -5\text{ m/s}^2

.

3

Sum the reaction distance and the braking distance to find the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 15\text{ m} + 90\text{ m} = 105\text{ m}

The total stopping distance is the sum of the distance covered before braking begins and the distance covered while braking.

Key Concept

Two-phase linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Question 10Question →

A stone is projected vertically upwards from the top edge of a cliff $80\text{ m}$ high with an initial speed of $30\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the total time, in seconds, taken by the stone to reach the ground at the base of the cliff.

Show answer & explanation

Answer: 8

Answer

The total time taken by the stone to reach the ground at the base of the cliff is

8\text{ s}

.

Using the equation of motion

s = ut - \frac{1}{2}gt^2

with

s = -80\text{ m}

,

u = 30\text{ m/s}

, and

g = 10\text{ m/s}^2

yields the quadratic equation

t^2 - 6t - 16 = 0

. Solving gives

t = 8\text{ s}

(ignoring the unphysical negative root

t = -2\text{ s}

). Alternatively, breaking the motion into two parts: time to reach maximum height (

30\text{ m/s} / 10\text{ m/s}^2 = 3\text{ s}

, covering

45\text{ m}

) plus time to fall from maximum height of

125\text{ m}

to the ground (

t = \sqrt{2(125)/10} = 5\text{ s}

), giving a total time of

3 + 5 = 8\text{ s}

.

Step-by-Step Solution

1

Set up the kinematic equation with appropriate vector signs

Displacement

s = -80\text{ m}

, initial velocity

u = +30\text{ m/s}

, acceleration

a = -g = -10\text{ m/s}^2

Since the ground is below the release point, displacement is negative when taking the upward direction as positive.

2

Substitute values into

s = ut + \frac{1}{2}at^2

-80 = 30t - 5t^2

Relates displacement, initial speed, time, and constant gravitational acceleration.

3

Form and solve the quadratic equation

5t^2 - 30t - 80 = 0 \implies t^2 - 6t - 16 = 0 \implies (t - 8)(t + 2) = 0

Simplifies the algebraic expression to find the time roots.

4

Select the physical root

t = 8\text{ s}

Time elapsed must be a positive quantity.

Key Concept

Kinematics of Vertical Motion under Gravity with Displacement from Elevation

Estimated Time:2m 0s

Question 11Question →

An electric train starts from rest and accelerates uniformly at $2\text{ m/s}^2$ for a time interval $t$ . It then maintains the maximum velocity attained for a time interval of $t$ . Finally, it decelerates uniformly at $4\text{ m/s}^2$ until it comes to a complete stop. If the total distance covered during the entire journey is $350\text{ m}$ , what is the value of $t$ ?

5\text{ s}

7\text{ s}

10\text{ s}

14\text{ s}

Show answer & explanation

Answer: $10\text{ s}$

Answer

10\text{ s}

The motion consists of three stages: acceleration from rest (

s_1 = t^2

), uniform velocity (

s_2 = 2t^2

), and deceleration to rest (

s_3 = 0.5t^2

). Adding these gives a total distance of

S = 3.5t^2

. Setting

3.5t^2 = 350\text{ m}

yields

t^2 = 100

, so

t = 10\text{ s}

.

Step-by-Step Solution

1

Analyze Stage 1 (Acceleration phase)

Maximum velocity

v = 2t

, displacement

s_1 = \frac{1}{2}(2)t^2 = t^2

Starting from rest (

u = 0

) with acceleration

a_1 = 2\text{ m/s}^2

for duration

t

,

v = u + a_1 t = 2t

and

s_1 = \frac{1}{2} a_1 t^2 = t^2

.

2

Analyze Stage 2 (Constant velocity phase)

Displacement

s_2 = (2t)(t) = 2t^2

The train moves at constant velocity

v = 2t

for duration

t

, so distance is velocity multiplied by time.

3

Analyze Stage 3 (Deceleration phase)

Deceleration time

t_3 = 0.5t

, displacement

s_3 = 0.5t^2

Decelerating from

v = 2t

to

0

at

a_2 = 4\text{ m/s}^2

takes time

t_3 = \frac{v}{a_2} = \frac{2t}{4} = 0.5t

. Distance

s_3 = \frac{1}{2} v t_3 = \frac{1}{2} (2t)(0.5t) = 0.5t^2

.

4

Calculate total displacement and solve for

t

Total distance

S = 3.5t^2 = 350 \implies t^2 = 100 \implies t = 10\text{ s}

Summing the displacements:

S = t^2 + 2t^2 + 0.5t^2 = 3.5t^2

. Equating to

350\text{ m}

gives

3.5t^2 = 350

, so

t^2 = 100

, giving

t = 10\text{ s}

.

Key Concept

Multi-stage linear motion and displacement calculation

Estimated Time:2m 0s

Question 12Question →

A car initially traveling at a constant speed of $15\text{ m/s}$ accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$ . It then maintains the acquired maximum speed for $10\text{ s}$ before coming to rest under uniform retardation in $4\text{ s}$ . What is the total distance covered by the car during the entire motion?

Show answer & explanation

Answer: 400

Answer

The total distance covered by the car during the entire motion is

400\text{ m}

.

The total distance is calculated by summing the distances covered in the three distinct phases of motion: acceleration (

100\text{ m}

), uniform velocity (

250\text{ m}

), and uniform retardation (

50\text{ m}

), yielding a total distance of

400\text{ m}

.

Step-by-Step Solution

1

Calculate final speed and distance for the acceleration phase.

Final speed

v = 25\text{ m/s}

and distance

s_1 = 100\text{ m}

.

Using kinematic equations

v = u + a t_1 = 15 + (2)(5) = 25\text{ m/s}

and

s_1 = u t_1 + \frac{1}{2}a t_1^2 = 15(5) + \frac{1}{2}(2)(5^2) = 75 + 25 = 100\text{ m}

.

2

Calculate the distance covered during the constant speed phase.

Distance

s_2 = 250\text{ m}

.

The car maintains the acquired speed of

25\text{ m/s}

for

10\text{ s}

, giving

s_2 = v \cdot t_2 = 25 \times 10 = 250\text{ m}

.

3

Calculate the distance covered during the retardation phase.

Distance

s_3 = 50\text{ m}

.

Using average velocity for uniform retardation to rest:

s_3 = \frac{v + 0}{2} t_3 = \frac{25}{2} \times 4 = 50\text{ m}

.

4

Sum the distances from all three stages to determine total distance.

Total distance

S = 400\text{ m}

.

S = s_1 + s_2 + s_3 = 100 + 250 + 50 = 400\text{ m}

.

Key Concept

Multi-stage linear motion and equations of uniform acceleration

Question 13Question →

A traffic officer on a stationary motorcycle spots a car passing at a constant speed of $15\text{ m/s}$ . The officer immediately pursues the car, accelerating uniformly from rest at $4\text{ m/s}^2$ for $5\text{ s}$ , after which the motorcycle continues at the constant speed attained. How long after setting off does the motorcycle overtake the car?

10\text{ s}

7.5\text{ s}

5\text{ s}

15\text{ s}

Show answer & explanation

Answer: $10\text{ s}$

Answer

The total time required for the motorcycle to overtake the car is

10\text{ s}

.

During the first

5\text{ s}

, the motorcycle accelerates from rest at

4\text{ m/s}^2

to

20\text{ m/s}

, covering

50\text{ m}

. In that same interval, the car moving at

15\text{ m/s}

covers

75\text{ m}

, creating a

25\text{ m}

separation gap. Beyond

5\text{ s}

, the motorcycle maintains

20\text{ m/s}

and closes in on the car (

15\text{ m/s}

) at a relative speed of

5\text{ m/s}

. It takes

5\text{ s}

to close the

25\text{ m}

gap, making the total pursuit time

10\text{ s}

.

Step-by-Step Solution

1

Calculate the state of both vehicles at the end of the motorcycle's acceleration phase (

t_1 = 5\text{ s}

).

Motorcycle speed

v = u + at = 0 + 4(5) = 20\text{ m/s}

. Motorcycle distance

s_1 = \frac{1}{2} a t^2 = \frac{1}{2}(4)(5^2) = 50\text{ m}

. Car distance

s_{\text{car}} = v_{\text{car}} \times t = 15 \times 5 = 75\text{ m}

.

Determine the position gap and relative speed after the acceleration phase ends.

2

Determine the remaining distance gap and relative speed between the vehicles.

Separation gap

d = 75\text{ m} - 50\text{ m} = 25\text{ m}

. Relative speed

v_{\text{rel}} = 20\text{ m/s} - 15\text{ m/s} = 5\text{ m/s}

.

Both vehicles now move at constant velocities, so relative speed determines how quickly the gap closes.

3

Calculate the time taken in the second phase and sum for total time.

Phase 2 time

t_2 = \frac{d}{v_{\text{rel}}} = \frac{25}{5} = 5\text{ s}

. Total time

t = t_1 + t_2 = 5\text{ s} + 5\text{ s} = 10\text{ s}

.

The total duration of pursuit is the acceleration duration plus the constant velocity catch-up duration.

Key Concept

Multi-stage linear motion involving uniform acceleration followed by constant velocity.

Estimated Time:1m 30s

Question 14Question →

A stone is thrown vertically downwards from the top of a $60\text{ m}$ high tower with an initial speed of $5\text{ m/s}$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , calculate the time taken, in seconds, for the stone to reach the ground.

Show answer & explanation

Answer: 3

Answer

The stone takes

3\text{ s}

to reach the ground.

Applying the equation

h = ut + \frac{1}{2}gt^2

with

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

g = 10\text{ m/s}^2

yields

60 = 5t + 5t^2

. Dividing by

5

gives

t^2 + t - 12 = 0

, which factorizes into

(t + 4)(t - 3) = 0

. Rejecting

t = -4\text{ s}

leaves the correct time of

3\text{ s}

.

Step-by-Step Solution

1

Set up the vertical motion equation

Using

h = ut + \frac{1}{2}gt^2

, substitute

h = 60\text{ m}

,

u = 5\text{ m/s}

, and

g = 10\text{ m/s}^2

.

This equation directly relates displacement, initial speed, acceleration, and time.

2

Form and solve the quadratic equation for time

60 = 5t + 5t^2 \Rightarrow t^2 + t - 12 = 0 \Rightarrow (t - 3)(t + 4) = 0

, giving

t = 3\text{ s}

.

Time must be positive, so the physically meaningful solution is

t = 3\text{ s}

.

Key Concept

Vertical motion under gravity with non-zero initial downward velocity

Question 15Question →

A ball is thrown vertically upwards with an initial velocity of $30\text{ m/s}$ from the edge of a cliff that is $35\text{ m}$ above ground level. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the total time elapsed before the ball hits the ground?

3.0\text{ s}

4.0\text{ s}

7.0\text{ s}

5.0\text{ s}

Show answer & explanation

Answer: $7.0\text{ s}$

Answer

The total time elapsed before the ball hits the ground is

7.0\text{ s}

.

Using the displacement equation

s = ut - \frac{1}{2}gt^2

with downward taken as negative, the net displacement when the ball reaches the ground is

-35\text{ m}

. Setting up the equation:

-35 = 30t - 5t^2

, which simplifies to

t^2 - 6t - 7 = 0

. Factoring gives

(t - 7)(t + 1) = 0

, yielding

t = 7.0\text{ s}

.

Step-by-Step Solution

1

Calculate the time taken (

t_1

) to reach maximum height

t_1 = \frac{u}{g} = \frac{30}{10} = 3.0\text{ s}

At maximum height, the final vertical velocity is

0\text{ m/s}

.

2

Calculate the maximum height (

h_{max}

) reached above the cliff

h_{max} = \frac{u^2}{2g} = \frac{30^2}{2(10)} = 45\text{ m}

Using the kinematic equation

v^2 = u^2 - 2gh

.

3

Determine total height above ground and time (

t_2

) to fall to the ground

Total height

H = 35\text{ m} + 45\text{ m} = 80\text{ m}

.

t_2 = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(80)}{10}} = 4.0\text{ s}

The ball falls from rest from a peak height of

80\text{ m}

.

4

Calculate the total time of flight

t_{total} = t_1 + t_2 = 3.0\text{ s} + 4.0\text{ s} = 7.0\text{ s}

The complete journey consists of ascending to the peak and descending to the ground.

Key Concept

Kinematics of Vertical Motion Under Gravity

Estimated Time:1m 30s

Question 16Question →

A car is traveling along a straight horizontal road at a constant speed of $20\text{ m/s}$ . The driver suddenly spots an obstacle ahead and takes $0.5\text{ s}$ to react before applying the brakes. Once the brakes are applied, the car decelerates uniformly at a rate of $4\text{ m/s}^2$ until coming to a complete stop. What is the total distance traveled by the car from the moment the driver spots the obstacle until the car stops completely?

50\text{ m}

60\text{ m}

110\text{ m}

40\text{ m}

Show answer & explanation

Answer: $60\text{ m}$

Answer

The total distance traveled by the car is

60\text{ m}

.

The motion consists of two distinct stages: first, constant speed motion during the

0.5\text{ s}

reaction time giving

s_1 = 20 \times 0.5 = 10\text{ m}

; second, uniform deceleration from

20\text{ m/s}

to rest over

s_2 = \frac{u^2}{2a} = \frac{400}{8} = 50\text{ m}

. Adding both stages yields a total distance of

60\text{ m}

.

Step-by-Step Solution

1

Calculate the reaction distance traveled at constant speed before braking.

s_1 = v \times t = 20\text{ m/s} \times 0.5\text{ s} = 10\text{ m}

.

During the reaction time, acceleration is zero, so distance equals speed multiplied by time.

2

Calculate the braking distance using the third equation of motion.

Using

v^2 = u^2 + 2as

:

0 = (20)^2 + 2(-4)s_2 \implies 8s_2 = 400 \implies s_2 = 50\text{ m}

.

The car decelerates from

u = 20\text{ m/s}

to

v = 0\text{ m/s}

at

a = -4\text{ m/s}^2

.

3

Sum the reaction distance and the braking distance to obtain the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 10\text{ m} + 50\text{ m} = 60\text{ m}

.

Total distance is the sum of distances covered in both stages of motion.

Key Concept

Multi-stage linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Estimated Time:1m 30s

Question 17Question →

An electric train accelerates uniformly from rest at a rate of $2\text{ m/s}^2$ and then immediately decelerates uniformly at $4\text{ m/s}^2$ until it comes to a complete stop. If the total distance covered by the train during this entire motion is $600\text{ m}$ , what is the total time taken for the journey?

20\text{ s}

25\text{ s}

30\text{ s}

40\text{ s}

Show answer & explanation

Answer: $30\text{ s}$

Answer

The total time taken for the journey is

30\text{ s}

.

For motion from rest back to rest, the velocity-time graph forms a triangle of total base

T

and height

v_{\text{max}}

. Acceleration time is

t_1 = v_{\text{max}}/2

and deceleration time is

t_2 = v_{\text{max}}/4 = 0.5 t_1

. The total displacement is

s = \frac{1}{2} (2) t_1^2 + \frac{1}{2} (4) t_2^2 = t_1^2 + 2 (0.5 t_1)^2 = 1.5 t_1^2 = 600\text{ m}

, yielding

t_1 = 20\text{ s}

and

t_2 = 10\text{ s}

. Summing these gives a total journey time of

30\text{ s}

.

Step-by-Step Solution

1

Relate maximum velocity to time in each stage.

Let

v_{\text{max}}

be the maximum velocity. Acceleration time

t_1 = \frac{v_{\text{max}}}{2}

and deceleration time

t_2 = \frac{v_{\text{max}}}{4}

.

Using

v = u + at

from rest to

v_{\text{max}}

and from

v_{\text{max}}

to rest.

2

Express total time

T

in terms of maximum velocity.

T = t_1 + t_2 = \frac{v_{\text{max}}}{2} + \frac{v_{\text{max}}}{4} = \frac{3}{4} v_{\text{max}}

, which gives

v_{\text{max}} = \frac{4}{3} T

.

Total duration is the sum of individual phase durations.

3

Set up distance equation from the area of the velocity-time graph.

\text{Total distance } s = \frac{1}{2} \times T \times v_{\text{max}} = \frac{1}{2} \times T \times \frac{4}{3} T = \frac{2}{3} T^2 = 600\text{ m}.

Area under a velocity-time triangle equals total displacement.

4

Solve for total time

T

.

T^2 = 600 \times \frac{3}{2} = 900 \implies T = 30\text{ s}

.

Taking the square root gives the total travel time.

Key Concept

Multi-stage linear motion with uniform acceleration and deceleration

Question 18Question →

An electric scooter starts from rest and accelerates uniformly at a rate of $3\text{ m/s}^2$ for a time of $6\text{ s}$ . What is the total distance, in meters, traveled by the scooter during this period?

Show answer & explanation

Answer: 54

Answer

The total distance traveled by the scooter is

54\text{ m}

.

Using the second equation of linear motion

s = ut + \frac{1}{2}at^2

with

u = 0\text{ m/s}

,

a = 3\text{ m/s}^2

, and

t = 6\text{ s}

yields

s = 0 + \frac{1}{2}(3)(36) = 54\text{ m}

.

Step-by-Step Solution

1

Identify known variables from the stem

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 3\text{ m/s}^2

, time interval

t = 6\text{ s}

Extracting given values provides the foundation for selecting the correct kinematic equation.

2

Select the appropriate equation of linear motion

s = ut + \frac{1}{2}at^2

This formula directly relates displacement

s

to initial velocity

u

, constant acceleration

a

, and time

t

.

3

Calculate the numerical displacement

s = (0\text{ m/s})(6\text{ s}) + \frac{1}{2}(3\text{ m/s}^2)(6\text{ s})^2 = 0 + \frac{1}{2}(3)(36) = 54\text{ m}

Evaluating the mathematical expression yields the final total distance.

Key Concept

Uniform Linear Acceleration

Question 19Question →

A motorcycle traveling at a constant speed of $18\text{ m/s}$ passes a landmark. Exactly $4.0\text{ s}$ after passing the landmark, the rider accelerates uniformly at a rate of $2.5\text{ m/s}^2$ until reaching a speed of $28\text{ m/s}$ . What is the total distance, in meters, traveled by the motorcycle from the instant it passed the landmark to the moment it reaches $28\text{ m/s}$ ?

Show answer & explanation

Answer: 164

Answer

The total distance traveled by the motorcycle is

164\text{ m}

.

The total distance is obtained by finding the displacement during the

4.0\text{ s}

of constant speed at

18\text{ m/s}

(

72\text{ m}

) and adding the displacement during uniform acceleration from

18\text{ m/s}

to

28\text{ m/s}

at

2.5\text{ m/s}^2

(

92\text{ m}

), giving a sum of

164\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered during the initial constant speed stage.

The distance covered in the first

4.0\text{ s}

is

72\text{ m}

.

At a constant velocity

v = 18\text{ m/s}

, the distance

s_1 = v \times t = 18 \times 4.0 = 72\text{ m}

.

2

Calculate the distance covered during the accelerated motion stage.

The distance covered while accelerating from

18\text{ m/s}

to

28\text{ m/s}

is

92\text{ m}

.

Using the kinematic equation

v^2 = u^2 + 2as_2

, substitute

u = 18\text{ m/s}

,

v = 28\text{ m/s}

, and

a = 2.5\text{ m/s}^2

to get

28^2 = 18^2 + 2(2.5)s_2 \implies 784 = 324 + 5s_2 \implies s_2 = 92\text{ m}

.

3

Find the total distance traveled.

Total distance

s_{\text{total}} = 164\text{ m}

.

The total distance is the sum of the distance covered during the constant speed period (

72\text{ m}

) and during uniform acceleration (

92\text{ m}

).

Key Concept

Kinematics equations for multi-stage motion combining uniform speed and uniform acceleration.

Question 20Question →

A traffic officer on a stationary motorcycle spots a car moving past at a constant speed of $30\text{ m/s}$ . The officer takes $2\text{ s}$ of reaction time before accelerating uniformly at $4\text{ m/s}^2$ to a maximum cruise speed of $40\text{ m/s}$ , after which the motorcycle continues at this constant speed. What is the total distance traveled by the motorcycle from its initial position to the point where it catches up with the car?

360\text{ m}

510\text{ m}

600\text{ m}

840\text{ m}

Show answer & explanation

Answer: $840\text{ m}$

Answer

840\text{ m}

The correct answer is

840\text{ m}

. Over the first

2\text{ s}

, the motorcycle is stationary while the car covers

60\text{ m}

. Over the next

10\text{ s}

, the motorcycle accelerates to

40\text{ m/s}

, covering

200\text{ m}

, while the car travels another

300\text{ m}

(totaling

360\text{ m}

). To close the remaining

160\text{ m}

gap at a relative speed of

10\text{ m/s}

requires an additional

16\text{ s}

. The total elapsed time of

28\text{ s}

yields a total catch-up distance of

30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

Step-by-Step Solution

1

Calculate car position and motorcycle position at the end of the reaction time period (

t = 2\text{ s}

).

During the reaction time

t_r = 2\text{ s}

, the motorcycle remains stationary (

s_m = 0\text{ m}

). The car travels a distance

s_c = 30\text{ m/s} \times 2\text{ s} = 60\text{ m}

.

The motorcycle does not begin accelerating until after the officer's reaction time elapses.

2

Determine the time and distance required for the motorcycle to reach its maximum speed of

40\text{ m/s}

.

Time to reach maximum speed:

t_{acc} = \frac{v_{max} - u}{a} = \frac{40 - 0}{4} = 10\text{ s}

. Distance during acceleration:

s_{acc} = \frac{v_{max}^2 - u^2}{2a} = \frac{40^2 - 0}{2(4)} = 200\text{ m}

.

The motorcycle accelerates uniformly from rest until reaching its capped maximum velocity.

3

Calculate total positions at

t_{total1} = 2\text{ s} + 10\text{ s} = 12\text{ s}

from the instant the car passed.

Motorcycle position:

x_m(12) = 200\text{ m}

. Car position:

x_c(12) = 30\text{ m/s} \times 12\text{ s} = 360\text{ m}

. Distance gap remaining:

\Delta x = 360\text{ m} - 200\text{ m} = 160\text{ m}

.

Comparing positions at

12\text{ s}

establishes the remaining distance gap to be closed during the constant speed phase.

4

Calculate the time required during the constant-speed phase to close the remaining gap and find the total meeting distance.

Relative speed:

v_{rel} = 40\text{ m/s} - 30\text{ m/s} = 10\text{ m/s}

. Additional time needed:

\Delta t = \frac{160\text{ m}}{10\text{ m/s}} = 16\text{ s}

. Total elapsed time:

t = 12\text{ s} + 16\text{ s} = 28\text{ s}

. Catch-up distance:

s_{total} = 30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

With both vehicles moving at constant speeds, the relative velocity determines how quickly the remaining separation is eliminated.

Key Concept

Multi-stage linear motion involving delayed reaction time, uniform acceleration, and constant speed pursuit.

Estimated Time:3m 0s

Kinematics and Linear Motion

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