A race car is moving with a velocity of 144 kilometers/hour. The driver applies the brakes, and the car comes to a halt in 12.0 seconds. What is the acceleration of the car after 12.0 seconds?
a) -12.0 meters/second^2
b) -3.3 meters/second^2
c) 3.3 meters/second^2
d) 10.3 meters/second^2

Question

A race car is moving with a velocity of 144 kilometers/hour. The driver applies the brakes, and the car comes to a halt in 12.0 seconds. What is the acceleration of the car after 12.0 seconds?
a)  -12.0 meters/second^2
b) -3.3 meters/second^2
c) 3.3 meters/second^2
d) 10.3 meters/second^2

theeric · Answer

This question is worded badly.. But, it wants the average acceleration while breaking. Do you have any ideas about what you'll need to be doing in this problem?

anonymous · Answer

is it distance subtracted by time, then divided by the time?

theeric · Answer

Well, you don't have the distance there... You don't need it anyway! But I'm glad you're thinking!

You need to know what acceleration is. Would you like me to tell you, or do you want to tell me what it is?

anonymous · Answer

i divided 144 by the 12.0 and got 12 as my answer, is that correct?

theeric · Answer

Nope, and I'll tell you why!

First of all, you have the right idea. Acceleration is the change in velocity per the change in time. But, $\dfrac{144\ [km\ /\ h]}{12\ [s]}=12\ [km\ s\ /\ h]$. Units is the problem, there.

There is also the issue of direction! The change in velocity is the final amount minus the initial amount. That is $0-144=-144$. Think, that really is the change. See, you're at 144km/h. How do you get to 0? You change -144 to get to 0.

theeric · Answer

So, let's fix that first problem. Change the units! Do you have experience with converting things?

anonymous · Answer

not really this isnt my homework

theeric · Answer

You're asking this for someone else, you mean?

anonymous · Answer

yes

theeric · Answer

If so, here's what you can show the person who needs to do it.

$\bar a=\dfrac{v_f-v_i}{\Delta t}$

$1\ [km]=1,000\ [m]\implies 1=\dfrac{1,000\ [m]}{1\ [km]}$

$1\ [hr]=3,600\ [s]\implies 1= \dfrac{1\ [hr]}{3,600\ [s]}$

$144\ [km/hr]\times1\times1\=144\ [km/hr]\times\dfrac{1,000\ [m]}{1\ [km]}\times  \dfrac{1\ [hr]}{3,600\ [s]}=40\ [m/s]$

So, your change is $-40\ [m/s]$, and the time interval is $12s$. Now you can use division like you tried before. That is because $\bar a=\dfrac{v_f-v_i}{\Delta t}$.