Question

A runner starts from rest and stops in 12 seconds. He covers

Answer 1

Answer:

b 1.39 m/s²

Explanation:

Given the following data;

Time = 12 seconds

Distance, S = 100 m

Since it's starting from rest, the initial velocity is equal to 0m/s.

To find the acceleration, we would use the second equation of motion;

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

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 0(12) + ½*a*12²

100 = 0 + 72

100 = 72a

Acceleration, a = 100/72

Acceleration, a = 1.389 ≈ 1.39 m/s²