# A 2230 kg car traveling to the west at 17.5 m/s slows down uniformly. How long would it take the car to come to a stop if the force on the car is 8810 N to the east? Let East be positive. Answer in...

A 2230 kg car traveling to the west at 17.5 m/s slows down uniformly.

How long would it take the car to come to a stop if the force on the car is 8810 N to the east? Let East be positive.

What is the car’s displacement during the time it takes to stop?

The answer to the first question is 4.429s, but I am not sure how to answer the second question. Any help would be appreciated. Thank you.

llltkl | Student

A 2230 kg car traveling to the west at 17.5 m/s slows down uniformly under the influence of a force of 8810 N to the east.

Let the uniform acceleratioin of the car be a.

By balancing force in the direction of motion (East-west),

`m*a=F_(nt)`

i.e. `2230*a=8810`

`rArr a=8810/2230=3.951 m/s^2` to the east (positive).

Further let the time required to come to standstill under this force be t and displacement by the time it stiops be s.

Apply `v=u+at` to find the time required:

`0=-17.5+3.951*t`

`rArr t=17.5/3.951=4.429 s`

Now apply `v^2=u^2+2as` to find the displacement:

`0=(-17.5)^2+2*3.95*s`

`rArr s=(-306.25 (m/s)^2)/(2*3.951 m/s^2)`

`=-38.76 m` (negative value implies that displacement would be to the west)

Therefore, displacement of the car during the time it takes to stop is 38.76 m.

PHYSICS101 | Student

I figured it out. I forgot to make west negative and east positive. The final answer was -38.7592 m.