stoping distance

2009-10-09 7:05 am

A car and a train move together along straight, parallel paths with the same constant cruising speed v_0. At t = 0 the car driver notices a red light ahead and slows down with constant acceleration - a_0. 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_0 with constant acceleration a_0. During the same time interval, the train continues to travel at the constant speed v_0.
Express your answer in terms of v_0 and a_0.
1. t_1
2.How much time does it take for the car to accelerate from the full stop to its original cruising speed?
t_2
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v_0 again?
delta_D

回答 (1)

天同

2009-10-10 5:30 am

✔ 最佳答案

How much time does it take for the car to accelerate from the full stop to its original cruising speed?

Using equation of motion, v=u + a.t
where u = 0 m/s, v = v_0, a = a_0, t =?
v_0 = (a_0).t
i.e. t = (v_0)/(a_0)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v_0 again?
Since the car deccelerated and accelerated at the same magnitude, the distances travelled during periods of decceleration and acceleration are the same.
During the decceleration, using equation of motion: v^2=u^2 + 2.a.s
u = v_0, v = 0 m/s, a = -a_0, s =?
0 = (v_0)^2 - 2.(a_0).s
i.e. s = (v_0)^2/2.(a_0)
Hence, total distance travelled by car = 2 x [(v_0)^2/2.(a_0)]
= (v_0)^2/(a_0)
During this time, total distance travelled by train
= (v_0).[2.(v_0)/(a_0)] = 2(v_0)^2/(a_0)
Therefore, distance of the car behind the train
= 2[(v_0)^2/(a_0)] - [(v_0)^2/(a_0)] = (v_0)^2/(a_0)

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