Question

My rour eacher 2 points SerPSES 2 P033 A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 12.0 m/s. Then the truck travels for 36.0 s stopping the truck in a uniform manner in an additional 5.00 ed until the brakes are applied, a) How long is the truck in motion? ) What is the average velocity of the truck for the motion described? m/s Need Help?

Answer 1

kinematic equation:
final velocity = initial velocity + acceleration multiplied by time
v1 = v0 + at

12 m/s = 0 m/s + 2 m/s^2 (t)
t = 6 seconds

a) total time = 6 + 36 + 5 = 47 seconds

b) must solve for total distance and divide it by time.

d1 = v0t + 1/2 a * t^2
d1 = 0 + .5(2) * 6^2
d1 = 36 meters

d2 = vt
d2 = 36 * 12
d2 = 432 meters

v1 = vo + at
0 = 12 + a(5)
-12/5 = a
a = -2.4 m/s^2

d3 = v0t + 1/2 a * t^2
d3 = 12 (5) - .5(2.4)*25
d3 = 30 meters

total distance = d1 + d2 + d3 = 36 + 432 + 30 = 498 meters

Average velocity = displacement / time elapsed= D/T = 498m/47s=10.59m/s