Physics Question?- 10 POINTS FOR BEST ANSWER?

Question

Brett starts from rest on his skateboard and accelerates at 0.50 m/s^2 [N] for 12 s, and then maintains his velocity for 15 s before falling off his board. He slides along on his back until he comes to rest, accelerating at -2.4 m/s^2:

(a) Calculate the highest instantaneous velocity that Brett reaches.

(b) How long did it take Brett to come to rest after falling?

(c) What is his displacement in the first 12 s?

(d) What is his displacement after he fell?

(e) WHat is Brett's average velocity over the total time?

Answer 1

Every best answer gets 10 points (plus the thumbs up) silly

Answer 2

a.

Highest velocity is at point where he stops accelerating:

v = a*t = 0.5*12 = 6m/s

b.

v = u + at

0 = 6 -2.4*t

t = 6/2.4 = 2.5 s

c.

s = ut + 0.5at^2

= 0 + 0.5*0.5*12^2

= 36m

d.

After he fell but before he came to rest displacement = 36m + distance travelled at constant velocity.

s = 6*15 = 90m

Total = 126 m

The question is not clear.

If the question means when he came to rest you need to add on the distance travelled after he fell.

v^2 = u^2 + 2as

0 = 6^2 + 2*(-2.4)*s

= 36 - 4.8s

s = 36/4.8 = 7.5m

So total distance = 126 + 7.5 = 133.5 m

e.

Average velocity: total distance/ total time

= 133.5/(12 + 15 + 2.5)

= 133.5/29.5 = 4.525 m/s