A particle starts from rest and accelerates uniformly at a rate of 4\text{ m/s}^2 for a du… — JAMB UTME

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?

Answer: 30 s

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

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

Relate deceleration time

t_2

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

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

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}

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

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