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

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in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..