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just olya [345]
3 years ago
11

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

driver notices a red light ahead and slows down with constant acceleration -a(initial). Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v(initial) with constant acceleration a(initial). During the same time interval, the train continues to travel at the constant speed v(initial).
How much time does it take for the car to come to a full stop? t1 = ?
How much time does it take for the car to accelerate from the full stop to its original speed? t2 = ?
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v(initial) again?
delta d = ?
Physics
1 answer:
kogti [31]3 years ago
8 0

Answer:

t_1 = \frac{v_i}{a_i}

t_2 = \frac{v_i}{a_i}

Δd = v_it_1 = v_i^2/a_i

Explanation:

As v(t) = v_i + at, when the car is making full stop, v(t_1) = 0 . a = -a_i . Therefore,

0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}

Apply the same formula above, with v(t_2) = v_i and a = a_i, and the car is starting from 0 speed,  we have

v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}

As s(t) = vt + \frac{at^2}{2}. After t = t_1 + t_2, the car would have traveled a distance of

s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}

Hence s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}

As t_1 = t_2 we can simplify s(t) = v_it_1

After t time, the train would have traveled a distance of s(t) = v_i(t_1 + t_2) = 2v_it_1

Therefore, Δd would be 2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i

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