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A motor car moving at a speed of 72km/h can not come to a stop in less than 3.0 s while for a truck this time interval is 5.0 s. On a higway the car is behind the truck both moving at 72km/h. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is 0.5s.

0 13 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a long answer type question as classified in NCERT Exemplar

    speed of car and truck = 72km/h = 72 (5/18) =20m/s

    V= u+at

    0=20+a (5) so a=-4m/s2

    But retarted acceleration will be v=u+at

    0=20+a (3)

    So a= 20 3 t - 0.5 -20/3m/s2

    We also need to consider human response time = 0.5 s

    V=u-at (for retarded motion)

    V= 20- 20 3 t - 0.5 ….1

    Vt=20-4t ….2

    From 1 and 2

    20-=20-4t

    After solving we get t= 5/4s

    Distance travelled by truck in time t, S=ut+1/2at2

    = 20 × 5 4 + 1 2 - 4 × 5 4 2 = 21.875 m

    To avoid the bump onto the truck car must maintain distance = 23.125-21.875=1.250m

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