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A race car starts from rest and travels east along a straight and level track. For the first 5.0 ss of the car's motion, the eastward component of the car's velocity is given by: vx(t)= (0.910m/s^3)t^2.What is the acceleration of the car when vx=12.0m/s?

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onyebuchinnaji

Answer:

The acceleration of the car is 7.85 m/s²

Explanation:

Given;

vx(t)= (0.910m/s³)t², given time traveled by the car 't' = 5.0 s

⇒To determine the velocity for 5 seconds, we substitute in 5.0 s for t

vx(5)= (0.910m/s³)(5s)²

       = (0.910m/s³)(25s²)

      vx = 22.75 m/s

⇒To determine the acceleration of the car when vx=12.0m/s

Acceleration is change in velocity per unit time

when vx=12.0m/s, our new equation becomes; 12 = (0.910m/s³)t²

Solving for t: t² = 12/0.91

                     t² = 13.187

                      t = √13.187 = 3.63 s

  Acceleration = Δv/Δt

                        [tex]= \frac{(22.75 - 12)m/s}{(5-3.63)s} = (\frac{10.75}{1.37}).\frac{m}{s^2}[/tex]

  Acceleration = 7.85 m/s²

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