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faust18 [17]
3 years ago
5

A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan

ding 180 m from where the front of the train started what will be the speed of the last car as it passes the worker? front of moving. what will 75 m y23 m/s
Physics
1 answer:
Lapatulllka [165]3 years ago
8 0

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

v^2 - v_0^2 = 2aswhere v = 23 m/s is the velocity of the car when it passes the worker, v_0 = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

23^2 - 0^2 = 2*a*255

510a = 529

a = 529 / 510 = 1.04 m/s^2

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then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

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t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

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