Question

a soccer player is running at 6m/s. he then stumbles over an opponents foot falling, and rolling to a stop. this took 4 seconds. what was his acceleration?

Answer 1

$a = \frac{(vf - vi)}{t} \\ a = \frac{0 - 6}{4} \\ a = - 1.5 \frac{m}{ {s}^{2} }$
top one is the formula
mid- i plug in the numbers
bottom- answer

Answer 2

Answer:

The acceleration of the soccer player is $1.5\ m/s^2$ and it is decelerating.

Explanation:

Given that,

Initial speed of a soccer player, u = 6 m/s

Finally, it stops, v = 0

Time, t = 4 s

We need to find the acceleration of a soccer player. We know that the rate of change of velocity is called acceleration of an object. It is given by :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-6}{4}\\\\a=-1.5\ m/s^2$

So, the acceleration of the soccer player is $1.5\ m/s^2$ and it is decelerating.