Physics question, Train reducing speed with constant acceleration?

Question

A train 350 m long is moving on a straight track with a speed of 80.4 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 16.9 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.

I need some step by step help. It seems easy, but I just can't figure out where I'm getting stuck.

Answer 1

EDITED >

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the answerer below is right! I just revisted and worked for simpler solution because some persons are not comfortable with graphs

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look at the last car>

when breaks applied at t=0, last car was at x = 350 m moving with (u=80.4).

> when last car (t=t) crossed the crossing, it travelled s = 350 m, and at this time its speed was (v=16.9)

last car will cover the (350) meter separation with average speed

v(average) = [80.4 + 16.9]/2 = 48.65 km/h = 13.514 meter/sec

time of blockage = time of travel of last car = 350/v(average)

t = 350/13.514 = 25.9 sec

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another way

crossing O >>> ( -16.9 km/h)

................ | ...................................... |

................ | ......... 350 .................... | <<< O last car (80.4 km/h)

if we assume that the moment (t=0) breaks were applied, crossing started (as if) moving towards the last car with (- 16.9) then blockage will last till 350 m separation is overtaken completely

this will happen with >. 2L = 2*350 m = 700 m distance apart

and > they travel thus by relative speed [80.4 - (- 16.9)] = 97.3

v (relative) = 97.3 km/h = 27.03 m/s

.................................... last car

| ............ 350 .............. |

..................................... | .............. 350 .................. |

.....................................  crossing

t (cross) = 2*350/27.03 = 25.9 s

Answer 2

Taking the end of the last car as your reference point,

when the brakes are applied it is still travelling at (80.4 / 3.6) = 22.3 m/s [since 1m/s = 3.6 km/h] and it is still 350m from the crossing.

Therefore,

initial speed, u = 22.3 m/s

final speed, v = 16.9 / 3.6 = 4.69 m/s

distance, s = 350 m

By sketching a speed-time graph, where it is a straight line with a negative gradient and starting from 22.3 m/s, you will get a trapezium.

The straight line will be from 22.3 to 4.69, with t being the height of the trapezium. [That means that the two parallel sides of the trapezium are 22.3 and 4.69.]

area of the trapezium = 1/2 X (22.3 + 4.69) t

350 = 13.51 t

t = 25.9 s (ans)

Since the last point of the train will take 25.9s to pass the crossing, it also means that the crossing will be blocked for 25.9s.