Question

An object which is initially at rest,accelerates at a rate of 4 m/s^2. Its final position is 6 m from its initial position at the end of that acceleration. For how much time did it accelerate?

Answer 1

$time \: taken \: = \sqrt{3 \: } seconds$

Please refer to the image attached.

Answer 2

Answer:

• Time for which object accelerated = 1.73 sec   [approx.]

Explanation:

Given:-

• Initial velocity of object, u = 0   [since, it is at rest]
• Acceleration of object, a = 4 m/s²
• Distance covered by object, s = 6 m   [since, it's final position is 6 m from its initial position at the end of acceleration]

To find:-

• Time for which object accelerated, a =?

Formula required:-

• Second equation of motion

         s = u t + 1/2 a t²

Where,

s = distance covered

u = Initial velocity

t = time

a = acceleration

Solution:-

Using the second equation of motion

→ s = u t + 1/2 a t²

→ 6 = 0 × t + 1/2 × 4 × t²

→ 6 = 0 + 2 t²

→ t² = 6 / 2

→ t² = 3

→ t = 1.73 sec   [approx.]

Therefore,

• Object accelerated for 1.73 seconds approximately.