happykiddo one year ago A speeder traveling at 42 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 3.66 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder? Both located at the same place and time.

1. happykiddo

I used the formula $S=1/2(a)(t)^2+ Vi(t)+Si$

2. happykiddo

Because of the same place and time I made cop=speeder

3. happykiddo

So the way I did it is (1/2)(3.66m/s^2)(t^2)=(42m/s)(t)

4. happykiddo

I got it!! I incorrectly punched some numbers into the calculator before. But thank you for your effort : )

5. anonymous

The speeder is traveling at constant velocity The policeman is traveling at constant acceleration. For constant velocity I would use $d=vt$ and for constant acceleration, I would use $final distance=initial distance+vt+\frac{ 1 }{ 2 }at^2$ Well, what do you know? They both equals to distance traveled. So you set them equal to each other by solving for time.

6. anonymous

-_- lol. Right when I almost finished.

7. happykiddo

But still thanks!! Could you help with another problem?

8. anonymous

Writing in the equation sheet usually delays me.

9. anonymous

Yea sure.

10. happykiddo

I'l lpost it as a new question.

11. anonymous

Ok I'll be there.

12. happykiddo

A car traveling at 16 m/s accelerates at 3.95 m/s2 for 6 seconds. To the nearest meter how far does it travel? I used the same equation from the last question but got 77.1 meters instead of 167(correct answer).

13. happykiddo

$S=\frac{ 1 }{ 2 }(a)(t)^2+Vi(t)+Si$

14. happykiddo

a=3.95 t=6 Vi=0 Si=0

15. anonymous

No, the initial velocity is 16m/s. It does not say the vehicle started from rest.

16. anonymous

16m/s*6s=96m

17. happykiddo

Gotcha, let me try it out.

18. happykiddo

I got it! Thank you for the help.

19. anonymous

No problem!