## anonymous 4 years ago A speeder passes a parked police car at a constant speed of 28.4 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.93 m/s2. How much time passes before the speeder is overtaken by the police car? Answer in units of s

1. JamesJ

Let's set x = 0 where the police car is. What's the equation of motion for the two cars? $x_{speeder}(t) = ... what?$ $x_{police}(t) = ... what?$

2. JamesJ

the speeder's location xs(t) is xs(t) = 28.4t Now what's the police car's position, xp(t) ?

3. anonymous

(1/2)(2.93)t^2

4. JamesJ

Right, exactly. Now when are the two equal?

5. anonymous

set them = to each other?

6. anonymous

when the police car catches the speeder

7. JamesJ

Set them equal and find the value of t for which that is the case.

8. anonymous

48.46 and 0

9. anonymous

obviously its not 0

10. JamesJ

Right. t = 0 is when the driver first goes past the police car. You want the other answer.

11. anonymous

and it is not correct. I've already tried it previously.

12. JamesJ

t = 48 sec isn't right. Check your calculation

13. JamesJ

$28.4 t = \frac{2.93}{2} t^2$

14. JamesJ

$t = 0$ or $t = \frac{2 \times 28.4}{2.93}$

15. anonymous

I made an error somewhere, thank-you.

16. anonymous

2.93? wait

17. anonymous

2 * a a=.586?

18. anonymous

.586^2 ?

19. anonymous

noooo

20. anonymous

1/2 * 2.93?

21. JamesJ

The position of the speeder is$x_s(t) = 28.4 t$The position of the police car is $x_p(t) = \frac{1}{2} 2.93t^2$ Hence the positions are equal when $28.4 t = \frac{1}{2}2.93t^2$ Now solve for t as I've written above. You can factor out a t from both sides if t is not zero, which gives you $\frac{1}{2}2.93t = 28.4$ hence $t = \frac{2 \times 28.4}{2.93}$ Now simplify that expression.

22. anonymous

19.39

23. JamesJ

Yes, t = 19.39 seconds

24. anonymous

I was making this more difficult by using quadratic equation. Thank you for the help and clarification.