Question

A student drives by a stationary police car at 41.7m/s, exactly 7s later the policeman starts after the speeding car, accelerating at a steady 3.75m/s2. How long after the police car begins moving does it take to catch up with the speeding car.
i got 34.8s but its wrong, can anyone explain it.
Xcar= 291.9+41.7t
Xpolice= 1.875t2
Xcar=X police 1.875t2-41.7t-291.2
t=34.8s

Answer 1

equation i used was Xf= Xi +Vit+ (1/2)at^2

Answer 2

Is 291.9 the distance the car travels in seven seconds?

Answer 3

yea

Answer 4

Looks good. Double check that you used the correct solution to the quadratic.

Answer 5

Maybe its just a typo in the solution manual......it has 23.8s

Answer 6

The math gods seem to agree. http://www.wolframalpha.com/input/?i=0.5*3.75*%28t-7%29%5E2+%3D+41.7*t

Answer 7

I get 27.833 s.

Answer 8

@mathmate did you use the same equation?

Answer 9

I used your equation (corrected) and got 27.833, which is the same answer as eashmore's answer by subtracting 7 seconds.

Answer 10

Can you double check the data (speed, acceleration, 7 seconds, etc.)

Answer 11

Note that the clock starts when the police car moves, which is t=7 in eashmore's equation.

Answer 12

yea data is correct.......since the police car moves 7 seconds later i have cars position as t+7

Answer 13

Here is a better representation : http://www.wolframalpha.com/input/?i=0.5*3.75*%28t%29%5E2+%3D+41.7*%28t%2B7%29

Answer 14

ok, thanks a lot

Answer 15

Guess the answer key has a typo!