# Need help with an integration word problem. This appears to be unsolvable due to lack of information.

I'm not sure I understand what to do with what's given to me to solve this. I know it has to do with the relationship between velocity, acceleration and time.

At a distance of 45m from a traffic light, a car traveling 15 m/sec is brought to a stop at a constant deceleration.

a. What is the value of deceleration?

b. How far has the car moved when its speed has been reduced to 3m/sec?

c. How many seconds would the car take to come to a full stop?

Can somebody give me some hints as to where I should start? All I know from reading this is that $v_0=15m$, and I have no idea what to do with the 45m distance. I can't tell if it starts to slow down when it gets to 45m from the light, or stops 45m from the light.

Edit:

I do know that since accelleration is the change in velocity over a change in time, $V(t)=\int a\ dt=at+C$, where $C=v_0$. Also, $S(t)=\int v_{0}+at\ dt=s_0+v_0t+\frac{1}{2}at^2$. But I don't see a time variable to plug in to get the answers I need... or am I missing something?

-

Hint: Constant acceleration means that the velocity $v(t)=v(0)+at$ where $a$ is the acceleration. The position is then $s(t)=s(0)+v(0)t+\frac 12 at^2$. You should be able to use these to answer the questions.
I wasn't thinking properly when I labeled this question as physics. Yes, it's a physics type question, but this is a calculus class, and I'm asked to do this as an exercise in integration. I do know though, that since accelleration is the change in velocity over a change in time, so $\int a\ dt = at+c$, where $c=v_{0}$. Also, $\int v_{0}+at\ dt = s_{0} + v_{0}t + \frac{at^{2}}{2}$. But I don't see a time variable to plug in to get the answers I need... or am I missing something? – agent154 Jan 24 '13 at 18:33
@agent154: To find the time when the velocity is zero, use the velocity equation. Set $v(t)=0$ to get $t=\frac {-v_0}a$, then put this into the $s$ equation. You want $s(t)=45$, which will let you find $a,t$ – Ross Millikan Jan 24 '13 at 18:45