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car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver notices a red light ahead and slows down with constant acceleration −a0. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v0 with constant acceleration a0. During the same time interval, the train continues to travel at the constant speed v0.

Part A :- How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Part B:- How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

Part C:- The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

Respuesta :

Answer:

Part A: [tex]t = v_0/a_0[/tex]

Part B: [tex]t = v_0/a_0[/tex]

Part C: [tex]v_0^2/a_0[/tex]

Explanation:

Part A:

We will use the following kinematics equation:

[tex]v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}[/tex]

Part B:

We will use the same kinematics equation:

[tex]v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}[/tex]

Part C:

The total time takes is 2t.

So the train moves a distance of

[tex]x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}[/tex]

And the car moves a distance in Part A and in Part B:

[tex]d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}[/tex]

So the total distance that the car traveled is [tex]d = \frac{v_0^2}{a_0}[/tex]

The difference between the train and the car is

[tex]x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}[/tex]