A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the truck travels for 46.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s.

(a) How long is the truck in motion?

s

(b) What is the average velocity of the truck for the motion described?

m/s

total time = 65 seconds

total distance = 1554 meters

Explanation:

kinematic equation:

final velocity = initial velocity + acceleration multiplied by time

v_1 = v_0 + at

28 m/s = 0 m/s + 2 m/s^2 (t)

t = 14 seconds

a) total time = 14 + 46 + 5 = 65 seconds

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

d_1 = v_0t + 1/2 a * t^2

d_1 = 0 + 0.5 (2) * 14^2

d_1 = 196 meters

d2 = vt

d2 = 28 * 46

d2 = 1288 meters

v_1 = v_o + at

0 = 28 + a (5)

- 28/5 = a

a = - 5.6 m/s^2

d_3 = v_0t + 1/2 a * t^2

d_3 = 28 (5) - 0.5 (5.6) * 5^2

d_3 = 70 meters

total distance = d1 + d2 + d3 = 196 + 1288 + 70 = 1554 meters