Question

A car stops in 60m. It has an acceleration of -10m/s^2. How long did it take to stop?

Answer 1

Hi!

Answer: 3.45 seconds

First you note down the known values:

s= 60m

v=0

t= ?

a= -10m/s^2

The equation of motion linking these values in a manner as to obtain the unknown value of t is:

s = vt - 1/2at^2

rearranging the equation for time,

t^2 = ( vt - s ) x 2 / a

t^2 = (0t - 60) x 2 / -10m/s^2

t^2 = 12

t = 3.45 s

Answer 2

Answer:

3.46 seconds

Explanation:

We are given that a car stops in 60 m and has an acceleration of -10m/s$^2$ and we are to find the time it took to stop.

To find this time, we will use this equation of motion:

$s = vt - \frac {1} {2} a t^2$

Rearranging the formula by making t the subject and then substituting the given values in the above formula to find t:

$t^2=(vt-s)*\frac{2}{a}$

$t^2 = (0)t - 60 * \frac{2}{-10}$

$\sqrt{t}^2=\sqrt{12}$

$t=3.46$

Therefore, it took 3.46 seconds for the car to stop.