Question

Starting from rest, a sprinter reaches his top velocity in 3 seconds. He runs a distance of 24m in 3 seconds. What is his acceleration? (Assume his acceleration is uniform)

Answer 1

Answer:

$a=5.34\ m/s^2$

Explanation:

Given that,

Initial velocity of a sprinter, u = 0

He runs a distance of 24m in 3 seconds.

We need to find his acceleration. Let a be his acceleration. Using second equation of kinematics to find it.

$s=ut+\dfrac{1}{2}at^2\\\\24=0+\dfrac{1}{2}a(3)^2\\\\a=\dfrac{24\times 2}{9}\\\\a=5.34\ m/s^2$

So, the acceleration of the sprinter is $5.34\ m/s^2$.

Answer 2

The acceleration of the sprinter from rest to his top speed is 5.33m/s²,

Given the data in the question

Since the sprinter starts from rest,

• Initial velocity; $u = 0$

• Time taken; $t = 3s$

• Distance covered; $s = 24m$

Acceleration; $a = \ ?$

To determine the acceleration of the sprinter, we use the second equation of motion:

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

Where s is distance covered, u is initial velocity, a is acceleration and t is time.

We substitute our given values into the equation

$24m = [0 * 3s] + [\frac{1}{2} * a * (3s)^2]\\\\24m = \frac{1}{2} * a * 9s^2\\\\24m = a * 4.5s^2\\\\a = \frac{24m}{4.5s^2} \\\\a = 5.33m/s^2$

Therefore, the acceleration of the sprinter from rest to his top speed is 5.33m/s²

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