Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

-_- lol.
Right when I almost finished.

Answer 7

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

Answer 8

Writing in the equation sheet usually delays me.

Answer 9

Yea sure.

Answer 10

I'l lpost it as a new question.

Answer 11

Ok I'll be there.

Answer 12

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).

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

16m/s*6s=96m

Answer 17

Gotcha, let me try it out.

Answer 18

I got it! Thank you for the help.

Answer 19

No problem!