# Physics

posted by Isaac

A truck on a straight road starts from rest and accelerates at 2.8 m/s^2 until it reaches a speed of 16 m/s. Then the truck travels for 16s at constant speed until the brakes are applied,stopping the truck in a uniform manner in an additional 3.6 s.
How long is the truck in motion?

What is the average velocity of the truck for the motion described?

So I've done the first part already which is 25.31 s, but when i try to find the average velocity for part two. I keep getting it wrong...can anyone show me the process on how to correctly do the second one?

1. Henry

V = Vo + a*t = 16 m/s.
0 + 2.8*t = 16
t1 = 5.71 s. to reach 16 m/s.

d1 = 0.5a*t^2 = 0.5*2.8*5.71^2 = 45.65 m

d2 = V*t2 = 16 * 16 = 256 m.

t1+t2+t3 = 5.71 + 16 + 3.6 = 25.31 s in
motion.

V = Vo + a*t = 0
16 + a*3.6 = 0
3.6a = -16
a = -4.44 m/s^2.

d3 = (V^2-V0^2)/2a = (0-(16^2)/-8.88 =
28.8 m. To stop.

Vavg = (d1+d2+D3)/(t1+t2+t3) =
(45.65+256+28.8)/25.31 = 13.06 m/s.

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