A body accelerates uniformly from rest at a rate of 4\text{ m/s}^2 for 6\text{ s}. What is… — JAMB UTME

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?

$72\text{ m}$ Answer
B
$144\text{ m}$
C
$24\text{ m}$
D
$12\text{ m}$

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

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.

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.

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

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