Question

The train's machinist was told that there was an obstacle on the tracks and he would stop before reaching it. When

starts braking, machinist looks the speed radar and reads the speed 144 km/h. Deceleration is uniform and the train
stopped ahead the obstacle after 60 seconds.
Calculate the train's acceleration in those 60 seconds.
At what distance from train was the obstacle?

`clues \\ v0 = 144km.h \\ v = 0 \\ t = 60s \\ a = x \\ d = x \\ a)a = (v - va)/(t) \\ b)d = v0 * t + (at ^(2) )/(2)`
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Answer 1

Explanation: Here we have uniform acceleration there for we can se motion equation to find the acceleration and the distance

V = U + at

V = 0 when it wil be stopped

144 km/h = 40 m/ s

When we use motion equation we have to put all values in SI units

0 = 40 - a ×60

a = 0.67m/s²

And for calculating distance

S = (V+U)×t / 2

S = ( 0+40)60/2

S = 1200 m = 1.2 km