Kinematics and Linear Motion Alıştırma Soruları — JAMB UTME — Sayfa 2

Soru 21Soru →

A research rocket is launched vertically upwards from rest with a constant acceleration of $5.0\text{ m/s}^2$ . At an altitude of $250\text{ m}$ , its engine suddenly fails and the rocket continues to move vertically upward under gravity alone. Calculate the total time, in seconds, taken by the rocket from launch until it reaches its maximum height. (Take acceleration due to gravity $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

The total time taken from launch to reach maximum height is

15\text{ s}

.

The motion occurs in two phases. In phase 1, accelerating uniformly from rest at

5.0\text{ m/s}^2

over

250\text{ m}

yields a velocity of

50\text{ m/s}

in

10\text{ s}

. In phase 2, moving upward under gravity alone (

10\text{ m/s}^2

) reduces the velocity from

50\text{ m/s}

to rest (

0\text{ m/s}

) in

5\text{ s}

. Adding the durations of both phases gives

10\text{ s} + 5\text{ s} = 15\text{ s}

.

Adım Adım Çözüm

1

Calculate the rocket's velocity and elapsed time at the moment of engine failure.

Velocity

v_1 = 50\text{ m/s}

and time

t_1 = 10\text{ s}

.

The rocket accelerates uniformly from rest at

5.0\text{ m/s}^2

over a distance of

250\text{ m}

.

2

Calculate the duration of the unpowered upward motion until vertical velocity becomes zero.

Unpowered flight time

t_2 = 5\text{ s}

.

After engine failure, the rocket acts as a free projectile moving upward against gravity (

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

) with an initial velocity of

50\text{ m/s}

.

3

Sum the time intervals of both stages.

Total time

t_{\text{total}} = 10\text{ s} + 5\text{ s} = 15\text{ s}

.

The total motion consists of two distinct stages: powered acceleration followed by gravitational deceleration.

Anahtar Kavram

Multi-stage vertical motion under constant acceleration followed by free-fall under gravity

Soru 22Soru →

A body starting from rest accelerates uniformly along a straight path at a rate of $2.5\text{ m/s}^2$ . What is the distance covered by the body in $4.0\text{ s}$ ?

10\text{ m}

20\text{ m}

40\text{ m}

5.0\text{ m}

Cevabı ve açıklamayı göster

Cevap: $20\text{ m}$

Cevap

The distance covered by the body in

4.0\text{ s}

is

20\text{ m}

.

Using the kinematic relation

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

with

u = 0\text{ m/s}

,

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

, and

t = 4.0\text{ s}

gives

s = 0 + \frac{1}{2} \times 2.5 \times (4.0)^2 = 20\text{ m}

.

Adım Adım Çözüm

1

Identify the given kinematic values.

Initial velocity

u = 0\text{ m/s}

, acceleration

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

, and time

t = 4.0\text{ s}

.

The problem states the body starts from rest and undergoes uniform linear acceleration.

2

Select the appropriate formula for distance under uniform acceleration.

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

This equation connects initial velocity, acceleration, elapsed time, and total displacement.

3

Substitute the given values into the equation and compute displacement.

s = (0)(4.0) + \frac{1}{2}(2.5)(4.0)^2 = 0.5 \times 2.5 \times 16 = 20\text{ m}

.

Evaluating the expressions yields the distance traveled in meters.

Anahtar Kavram

Kinematic Equation for Linear Distance Under Uniform Acceleration

Tahmini Süre:45s

Soru 23Soru →

An electric train traveling along a straight track uniformly slows down from a speed of $30\text{ m/s}$ to $10\text{ m/s}$ over a distance of $200\text{ m}$ . What is the magnitude of the deceleration of the train in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The magnitude of the deceleration is

2\text{ m/s}^2

.

Using the kinematic equation

v^2 = u^2 + 2as

, substituting

v = 10\text{ m/s}

,

u = 30\text{ m/s}

, and

s = 200\text{ m}

gives

100 = 900 + 400a

, which simplifies to

400a = -800

, yielding

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

. The magnitude of deceleration is

2\text{ m/s}^2

.

Adım Adım Çözüm

1

Identify the given kinematic variables.

Initial velocity

u = 30\text{ m/s}

, final velocity

v = 10\text{ m/s}

, displacement

s = 200\text{ m}

.

Choosing the appropriate equation of motion requires knowing which variables are given and which is unknown.

2

Apply the third equation of motion relating initial velocity, final velocity, acceleration, and distance.

v^2 = u^2 + 2as

This formula connects

u

,

v

,

a

, and

s

without needing time

t

.

3

Substitute the given values into the equation and solve for acceleration

a

.

(10)^2 = (30)^2 + 2(a)(200) \implies 100 = 900 + 400a \implies 400a = -800 \implies a = -2\text{ m/s}^2

.

Performing algebraic operations to isolate the acceleration parameter.

4

State the magnitude of the deceleration.

The magnitude of deceleration is

2\text{ m/s}^2

.

Deceleration represents the rate of speed reduction, which corresponds to the magnitude of negative acceleration.

Anahtar Kavram

Uniformly Accelerated Motion Equations

Soru 24Soru →

Car $A$ , traveling at a constant speed of $20\text{ m/s}$ along a straight horizontal road, passes a landmark $50\text{ m}$ ahead of car $B$ , which is initially stationary. At $t = 0\text{ s}$ , car $B$ starts moving in the same direction, accelerating uniformly at $3.0\text{ m/s}^2$ until it reaches a top speed of $30\text{ m/s}$ , after which it continues at this constant top speed. How many seconds after $t = 0\text{ s}$ does car $B$ catch up with car $A$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

Car B catches up with car A after 20 seconds.

Car B accelerates for

10\text{ s}

covering

150\text{ m}

to reach

30\text{ m/s}

. During these

10\text{ s}

, car A reaches a position of

250\text{ m}

(taking into account its

50\text{ m}

head start). Car B then closes the remaining

100\text{ m}

gap at a relative speed of

10\text{ m/s}

(

30\text{ m/s} - 20\text{ m/s}

), taking an extra

10\text{ s}

. The total elapsed time is

10\text{ s} + 10\text{ s} = 20\text{ s}

.

Adım Adım Çözüm

1

Calculate the duration

t_1

of car B's acceleration phase to reach

30\text{ m/s}

.

t_1 = \frac{v_{max} - u}{a} = \frac{30 - 0}{3.0} = 10\text{ s}

Car B accelerates uniformly from rest at

3.0\text{ m/s}^2

until reaching its top speed limit.

2

Determine the distance

s_B

covered by car B and the position

s_A

of car A at

t = 10\text{ s}

.

s_B = \frac{1}{2}a t_1^2 = \frac{1}{2}(3.0)(10)^2 = 150\text{ m}

;

s_A = 50 + v_A t_1 = 50 + (20)(10) = 250\text{ m}

Car A starts

50\text{ m}

ahead and moves continuously at

20\text{ m/s}

.

3

Find the separation distance between the two cars at

t = 10\text{ s}

and compute the time

\Delta t

required to close it.

\text{Separation} = 250 - 150 = 100\text{ m}; \Delta t = \frac{100}{30 - 20} = 10\text{ s}

Beyond

t = 10\text{ s}

, car B travels at a constant relative velocity of

10\text{ m/s}

faster than car A.

4

Sum the acceleration time and constant speed time to find the total time taken.

t_{total} = t_1 + \Delta t = 10\text{ s} + 10\text{ s} = 20\text{ s}

Combining both phases yields the exact instant car B overtakes car A.

Anahtar Kavram

Multi-stage relative motion with acceleration limits

Soru 25Soru →

An electric cart moving along a straight horizontal track at an initial speed of $10\text{ m/s}$ accelerates uniformly at $3\text{ m/s}^2$ for a duration of $4\text{ s}$ . Immediately after this phase, it applies its brakes and decelerates uniformly at $2\text{ m/s}^2$ until it comes to a complete stop. What is the total distance, in meters, traveled by the cart during the entire motion?

Cevabı ve açıklamayı göster

Cevap: 185

Cevap

The total distance traveled by the cart during the entire motion is

185\text{ m}

.

The total distance is calculated by analyzing the two distinct stages of motion. In the first phase, traveling at an initial velocity of

10\text{ m/s}

with an acceleration of

3\text{ m/s}^2

for

4\text{ s}

yields a distance of

64\text{ m}

and a peak velocity of

22\text{ m/s}

. In the second phase, decelerating from

22\text{ m/s}

to rest at

2\text{ m/s}^2

requires a distance of

121\text{ m}

. Adding both distances (

64\text{ m} + 121\text{ m}

) yields the total distance of

185\text{ m}

.

Adım Adım Çözüm

1

Determine the distance (

s_1

) and final velocity (

v_1

) during the uniform acceleration phase

s_1 = 64\text{ m}

and

v_1 = 22\text{ m/s}

Using kinematic formulas

s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2

and

v_1 = u_1 + a_1 t_1

for constant acceleration.

2

Determine the stopping distance (

s_2

) during the uniform deceleration phase

s_2 = 121\text{ m}

Using

v_2^2 = v_1^2 - 2 a_2 s_2

with final speed

v_2 = 0\text{ m/s}

.

3

Sum the distances from both stages

S_{\text{total}} = 185\text{ m}

Total displacement for multi-stage motion along a straight line is the sum of displacements in each stage.

Anahtar Kavram

Multi-stage linear motion with constant acceleration and deceleration

Soru 26Soru →

An object is launched vertically upward from the ground with an initial velocity of $40\text{ m/s}$ . At the exact same instant, a second object is dropped from rest from a height of $100\text{ m}$ directly above the first object. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , at what time (in seconds) after launch will the two objects meet?

Cevabı ve açıklamayı göster

Cevap: 2.5

Cevap

The two objects meet after 2.5 seconds.

Because both objects experience identical downward gravitational acceleration (

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

), their relative acceleration is zero. The relative velocity between them remains constant at

40\text{ m/s}

. The time to cover the initial separation distance of

100\text{ m}

is calculated directly as

t = \frac{\text{distance}}{\text{relative velocity}} = \frac{100}{40} = 2.5\text{ s}

.

Adım Adım Çözüm

1

Set up the position-time equations for both objects taking ground level as

y = 0\text{ m}

.

For the upward-launched object:

y_1(t) = u t - \frac{1}{2}gt^2 = 40t - 5t^2

. For the dropped object:

y_2(t) = h_0 - \frac{1}{2}gt^2 = 100 - 5t^2

.

Kinematic equations of motion under uniform gravitational acceleration apply to both bodies.

2

Equate the two vertical position equations to solve for the meeting time

t

.

40t - 5t^2 = 100 - 5t^2 \implies 40t = 100

When the objects meet, they share the exact same vertical position

y_1(t) = y_2(t)

at time

t

.

3

Calculate the value of

t

.

t = \frac{100}{40} = 2.5\text{ s}

Direct algebraic division yields the time elapsed before collision/meeting.

Anahtar Kavram

Relative vertical motion under uniform gravity

Tahmini Süre:1m 30s

Soru 27Soru →

A particle traveling along a straight line accelerates uniformly from an initial speed $u$ to a final speed of $25\text{ m/s}$ over a distance of $150\text{ m}$ in a time duration of $8\text{ s}$ . What is the initial speed $u$ in $\text{m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 12.5

Cevap

The initial speed of the particle is

12.5\text{ m/s}

.

Under uniform acceleration, displacement is given by the product of average velocity and time:

s = \frac{u + v}{2}t

. Substituting

s = 150\text{ m}

,

v = 25\text{ m/s}

, and

t = 8\text{ s}

gives

150 = 4(u + 25)

, which yields

u = 12.5\text{ m/s}

.

Adım Adım Çözüm

1

Identify the given parameters and select the appropriate kinematic equation.

Givens:

s = 150\text{ m}

,

v = 25\text{ m/s}

,

t = 8\text{ s}

. Formula:

s = \frac{u + v}{2}t

.

This formula relates displacement directly to average velocity and time under uniform acceleration without needing the acceleration variable.

2

Substitute the given values into the formula.

150 = \left(\frac{u + 25}{2}\right) \times 8 = 4(u + 25)

.

Simplifying

\frac{8}{2}

gives a factor of

4

multiplying

(u + 25)

.

3

Isolate and solve for the unknown initial speed

u

.

u + 25 = \frac{150}{4} = 37.5 \implies u = 12.5\text{ m/s}

.

Subtracting

25

from

37.5

yields the value of

u

.

Anahtar Kavram

Kinematics with uniform linear acceleration

Kinematics and Linear Motion

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar