# Kinematic problems !

1. Sep 22, 2003

### x_endlessrain_x

Kinematic problems!!!!!!

hi there,here is my problem

The police observe that the skid marks of a stopping car are 200m long. Assuming the car decelerated at a constant rate of 20m/s^2,skidding all the way, how fast was the car going when the brakes were applied?

i think we can solve the problem by using the physics equation v^2=u^2+(2*(a*s)).......and
i try to use integration but the time interval isn't given so i am just looking for an alternative way to solve this problem.........by the way ,this is a calculus assignment.

thx

2. Sep 22, 2003

### Matt Jacques

Im getting Vi as 88.5 m/s

3. Sep 22, 2003

### x_endlessrain_x

yes i got 89 something too.......but i just wonder if this problem can be done by integration?

but thx anyway

Last edited: Sep 22, 2003
4. Sep 22, 2003

### Integral

Staff Emeritus
http://home.comcast.net/~rossgr1/Math/skidding.PDF [Broken] is a solution which uses integration. It starts with the basic fact that this is a problem with a constant acceleration and integrates the basic differential equation to arrive at the solution.

Last edited by a moderator: Apr 20, 2017
5. Sep 22, 2003

### x_endlessrain_x

thx integral it really help! :) thx a lot

6. Sep 23, 2003

### HallsofIvy

Staff Emeritus
Because this is a constant acceleration problem (so that the velocity function is linear), you can use the "average" velocity.

Let v be the initial speed. Of course, the ending speed is 0 so the "average" speed, the average of those two numbers, is v/2.

Letting t be the (unknown) time required to stop, the distance is
(v/2)t= 200 m.

And we do know the time- in a sense. At an acceleration of -20 m/s^2, to go from v to 0 requires (since 0= v- 20t) v/20 sec.

Putting that into (v/2)t= 200 gives v^2/40= 200 or v^2= 8000.

v= [sqrt](8000)= [sqrt](1600*5)= 40[sqrt](5)= 89.4 m/s

If you're not really sure about that "average" speed, here's the standard way:

With acceleration a, the distance travelled is d= (a/2)t^2+ vt.
(You could say we use integration to get that.)

In this problem, -10t^2+ vt= 200.

Again, since the time taken to slow from v to 0 with acceleration
-20 m/s^2 is v/20 sec., we have (-10)v^2/400+ v^2/20= 200.
That is, -v^2/40+ v^2/20= v^2/40= 200 so v^2= 8000 again.