A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 38.0 m/s. Then the vehicle moves for 85.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the ride-sharing car in motion (in s)?
(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)

Answer

case 1

initial speed of the car= 0 m/s

acceleration = 2 m/s²

final speed = 38 m/s

time taken to reach at the speed of 38 m/s

using equation of motion

v = u + at

38 = 0 + 2t₁

t₁ = 19 s

time for which car is moving with constant speed t₂ = 85 s

time taken to stop the vehicle t₃ = 5 s

a) Total ride time

    t = t₁ + t₂ + t₃

    t = 19 + 85 + 5

   t = 109 s

b) distance travel in case 1

  using equation of motion

   v² = u² + 2 a s

   38² = 2 x 2 x s₁

    s₁ = 361 m

  when car is moving with constant velocity

    s₂ = v t

    s₂ = 38 x 85

    s₂ = 3230 m

 when car is to stop

  using equation of motion to calculate the deceleration of the car

  v = u + at

  0 = 38 + a x 5

      a = -7.6 m/s²

  distance traveled

   v² = u² + 2 a s

   0² = 38² - 2 x 7.6 s₃

    s₃ = 95 m

total distance

    s = s₁ + s₂ + s₃

    s = 361 + 3230 + 95

    s = 3686 m

explanation:

newton's third law, for every action there is an equal and opposite reaction.

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