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21 August, 13:37

A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 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? (b) What is the average velocity of the truck for the motion described?

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  1. 21 August, 16:13
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    You can use the equation

    V_xf = V_xi + a_x (t)

    V_xf = 20.0m/s

    V_xi = 0m/s

    ax = 2.0 t

    Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

    a) 10 + 20 + 5 = 35 seconds

    for part b)

    You can use the formula

    Delta x/Delta t = average velocity

    Need to find xf, knowing xi = 0

    Thus, use the formula

    x_f = x_i + V_xi (t) + (1/2) a_x (t) ^ (2)

    x_f = 0 + 0 (10) + (1/2) (2.0) (10) ^ (2)

    x_f = 100m

    so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

    20m/s = xm/20s 20*20 = x

    x = 400

    thus we have 100+400 = 500m then it slows down from 500m to x_f

    thus I use the equation

    x_f = x_i + (1/2) (V_xf + V_xi) t

    x_f = 500 + (1/2) (0 + 20) (5) x_f = 500 + 50

    x_f = 550

    therefore the total distance traveled is 550m

    to calculate average velocity

    550/35 = 16m/s

    thus V_xavg = 16m/s
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