A body starting from rest accelerates uniformly along a straight path at a rate of 2.5\tex… — JAMB UTME

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 $4.0\text{ s}$ ?

A
$10\text{ m}$
$20\text{ m}$ Answer
C
$40\text{ m}$
D
$5.0\text{ m}$

Answer

The distance covered by the body in

4.0\text{ s}

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}

Step-by-Step Solution

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.

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.

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.

Key Concept

Kinematic Equation for Linear Distance Under Uniform Acceleration

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