How Far Does a Car Travel When Slowing Down?

[AymanLak]

Homework Statement
Initial velocity=15m/s slows down to 10m/s in 4.0s. Acceleration= -1.2m/s^2
How far does the car travel in 4.0s
If the car were to slow down at the same rate how much additional distance would it travel before coming to a complete stop

Homework Equations

A=d/t
D=(1/2)at^2

The Attempt at a Solution

I got the acceleration which was the first part of the question and so for the second part (calculating distance traveled in 4.0s) I plugged in the 1.25m/s^2 in the equation d=1/2at^2 and got 2.5m which was marked incorrect. For the third part I have no idea how to solve it

 

[collinsmark]

Hello AymanLak,

Welcome to Physics Forums! :)

Homework Statement
Initial velocity=15m/s slows down to 10m/s in 4.0s. Acceleration= -1.2m/s^2

I think you missed a key when typing in the "-1.2m/s^2". But I see that you corrected that below ($a = -1.25 \mathrm{\frac{m}{s^2}}$).

How far does the car travel in 4.0s
If the car were to slow down at the same rate how much additional distance would it travel before coming to a complete stop

Homework Equations

A=d/t
D=(1/2)at^2

The Attempt at a Solution

I got the acceleration which was the first part of the question and so for the second part (calculating distance traveled in 4.0s) I plugged in the 1.25m/s^2 in the equation d=1/2at^2 and got 2.5m which was marked incorrect. For the third part I have no idea how to solve it

There's two mistakes going on there.

(1) The uniform acceleration formula of $d = \frac{1}{2} a t^2$ only applies if the body starts from rest (i.e., if $v_0 = 0$). What is the full version of the formula if the object has a nonzero initial velocity?

(2) Don't forget to square the time in the $\frac{1}{2}a t^2$ term.