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?
Answer in units of s
What is the average velocity of the truck for the motion described?
Answer in units of m/s
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?
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
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.
posted by Henry