# 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.

Answer in units of s.

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

Answer in units of m.

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.

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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.**

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