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A motor car moving at a speed of 72 km/h can not come to a stop in less than 3 s while for a truck this time interval is 5.0s . On a highway the car is behind the truck both moving at 72 km/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 the truck . Human response time is 0.5s

  • physics -

    deceleration rate of the truck is

    at = 20 m/s / 5s = 4 m/s²

    deceleration rate of the car is

    at = 20 m/s / 3s = 6⅔ m/s²

    As the car can stop much quicker than the truck, then to avoid collision the car and truck velocities will be identical and non zero while the distance between them is near zero with a braking time (t) for the truck and (t - 0.5) for the car

    velocity of the truck
    vt = 20 - 4t

    velocity of the car
    vc = 20 - 6⅔(t - 0.5)
    vc = 20 - 6⅔t + 3⅓
    vc = 23⅓ - 6⅔t

    to find the time when both of these are true, set them equal

    23⅓ - 6⅔t = 20 - 4t
    3⅓ = 2⅔t
    t = 1.25 s

    In 1.25 seconds the car will have traveled a distance before braking occurs and a distance after braking starts

    dc = v₀t₀ + v₀t₁ + ½at₁²
    dc = 20(0.5) + 20(0.75) - ½(6⅔)(0.75)²
    dc = 23.125 m

    in 1.25 seconds, the truck will have traveled

    dt = 20(1.25) - ½(4)(1.25)²
    dt = 21.875 m

    so the minimum follow distance is

    23.125 - 21.875 = 1.25 m <=== ANSWER

    At 1.25 seconds, both vehicles will be traveling 15 m/s

    20 - 4(1.25) = 15
    20 - 6⅔(1.25 - 0.5) = 15

    For times greater than 1.25 seconds, the car is traveling slower than the truck and the gap between the car and truck increases.

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