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A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased ve
Question
A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 56mi/hr as it passes a point 1/2mi beyond A. A car moving at 45mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver “steps on the gas.” Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 3.9sec and find the velocity v of the car as it reaches the crossing.
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Physics
3 years
2021-08-01T06:54:50+00:00
2021-08-01T06:54:50+00:00 1 Answers
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Answer:
First to find deceleration of the train we use
v²= u²+ 2as
56²= 73²+ 2(0.5)a
a= -2193mi/hr²
Then we find time in which the train does the intersection
Using
S= ut+ 1/2 at²
1= 73t-1/2(1293)t²
t =68.5s
But since the train is to intersect in 3.9s the time will be the difference which is
65.68s
So finding acceleration
S= ut + 1/2at²
1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²
So a= 1.179ft/s²
To find velocity we use
V= u + at
= 45/3600mi/s + (-2.33E-4mis²)(65.58s)
V= 0.0271mi/s
= 97.6ft/s