How long is the truck in motion?What is the average velocity of the truck during the motion described?

May 13th, 2009 | by Michael |
truck
Victoria B asked:


A truck on a straight road starts from rest and accelerates at 2.0 m/s2 until it reaches a speed? 20 m / s. Then the truck travels for 20 s at speed? constant until the brakes are applied, stopping the truck in a uniform manner in an additional 5.0 S.
RICKEY

  1. 3 Responses to “How long is the truck in motion?What is the average velocity of the truck during the motion described?”

  2. By dudara on May 16, 2009 | Reply

    There are 3 parts to the motion

    part 1 takes 10 s

    part 2 takes 20 s

    and part 3 takes 5 s.

    Total time taken is 35 s

    To calculate the average velocity, we must work out the distance covered.

    Part 1: s = ut + 0.5*a*t^2

    s = 0 + 0.5*2*10^2

    s = 100 m

    Part 2: s = v*t = 20*20 = 400 m

    Part 3: First calculate the deceleration, then the distance

    v = u + at

    0 = 20 + a*5

    a = -4 m/s^2

    s = ut + 0.5*a*t^2

    s = 20*5 - 0.5*4*5^2 = 50m

    Total distance travelled is 550 m

    average velocity = distance/time = 550/35

    v = 15.7 m/s

  3. By pearlsawme on May 17, 2009 | Reply

    There are 3 parts to the motion

    part 1
    t 1=( v-u) /a
    t1 = 10 s

    part 2

    Given that t2 = 20s.
    part 3
    Given that t3 = 5 s.
    Total time taken is 35 s

    To calculate the average velocity; we must work out the total distance covered.

    Part 1:
    Average velocity is [20 +0] / 2 = 10
    s1 = 10 x 10 = 100 m

    Part 2:
    s2 = v*t = 20*20 = 400 m
    Part 3:
    Average velocity is [20 +0 ] / 2 = 10

    S3 = 10 x5 = 50 m

    Total distance traveled is 550 m

    average velocity = distance/time = 550/35

    v = 15.7 m/s

  4. By rana_22_m on May 17, 2009 | Reply

    first v=u+at where

    where v - final velocity a-acceleration t- time u- initial vel (zero in this case)
    since the rate of acceleration (a) is same till it reaches 20m/s

    time taken to reach 20m/s with an acceleration of m/s2 is 10 sec

    later it has travelled for 20 sec at constant speed and another 5 sec to come to rest

    over all it took 35sec travel.

    how far has it travelled??
    s=ut+0.5*a(t^2)
    as u = 0
    s=100m
    i.e b4 reaching 20m/s it has travelled 100m
    and at 20m/s it has travelled for 20 sec so tat makes it 400m
    aftet that it travels for 5 sec

    as v=u+at after that 5 sec v=0
    hence 0=20+a5
    thnce a=-4m/sec

    s(distance) = ut+0.5a (t^2)
    as u = 20m/s
    a= -4m/s2
    t=5
    s=50m

    hence it has travelled a total distance of 100+400+50 = 550m

    to calculate the average velocity it has travelled 550m in a total time of 35sec
    hence it is 550/35 = 15.71 m/s

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