Physics

posted by Neeraj

A car is travelling at 2 0m/s when it pulls out to pass a truck that is travelling at only 18 m/s the car accelerates at 2 m/s square for 4 sec and then maintains this new velocity. if the car was originally 8 mtrs behind the truck when it pulled out to pass , how far in front of the truck is the car 10 sec later

  1. Henry

    Vf = Vo + a*t = 20 + 2*4 = 28 m/s.
    Dc = 20*4 + 0.5*2*4^2 = 96 m. in 4 sec.
    Dc = 28m/s * 6s = 168 m. in 6 sec.
    Dc = 96 + 168 = 264 m. in 10 sec.

    Dt = 18m/s * 10s. = 180 m. in 10 sec.

    (264-8) - 180 = Distance in front of the truck.

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