The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a spe

Question

The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a speed 19.3m/s and its front bumper is a distance 25.0m behind the truck’s rear bumper. The car begins accelerating at a constant acceleration 0.560m/s^2 , then pulls back into the truck’s lane when the rear of the car is a distance 26.5m ahead of the front of the truck. The car is of length 4.50m and the truck is of length 20.7m .
Part A) How much time is required for the car to pass the truck?
Part B ) What distance does the car travel during this time?
Part C) What is the final speed of the car?

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MichaelMet 1 year 2021-08-31T22:22:44+00:00 1 Answers 34 views 0

Answers ( )

    0
    2021-08-31T22:23:53+00:00

    Answer:

    A)    t = 10.56 s, B)  x = 235 m, C) v = 25.2 m / s

    Explanation:

    A) We can solve this problem using kinematics expressions.

    The distance traveled by the truck is

           x_c = v_c t

    Distance traveled by the car.

    The car must travel the distance that separates them from the truck x₀=25.0.   Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2  4.5 = 9 m, therefore the total length traveled by the car is

              x_t = x₁ + x₂ + x₃

              x_t = 26.5 + 20.7 +9 = 56.2 m

    the distance traveled by the car when it returns to the lane is

             x_c + x_t = x₀ + v₀ t + ½ a t²

    when the car passes the car the distance traveled by the two vehicles is the same, we substitute

             v_c  t + x_t = x₀ + v₀ t + ½ a t²

             ½ a t² + t (v₀ -v_c) + (x₀ – x_t) = 0

    we substitute the values

             ½ 0.560 t² + t (19.3 -19.3) + (25.0 – 56.2) =

             0.28 t² -31.2 = 0

             t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]

             t = 10.56 s

    This is the time it takes for the car to pass the truck and back into the lane.

    B) the distance traveled is

            x = v₀ t + ½ a t²

            x = 19.3 10.56 + ½ 0.560 10.56²

            x = 235 m

    C) the final velocity is

             v = v₀ + a t

             v = 19.3 + 0.560 10.56

             v = 25.2 m / s

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