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 = ((vf - vi))/(t) \\ a = (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

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

Step-by-step 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=(v-u)/(t)\\\\a=(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.