A train is traveling at $90 k m / h .$ The brakes are applied so as to produce a uniform acceleration of $- 0.5 m / s^{2} .$ Find how far the train goes before it stops.

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Solution

Step 1: Given data
Acceleration of the train, $a = - 0.5 m / s^{2}$ .
Acceleration is negative as brakes are applied, therefore acceleration is acting in opposite direction of motion.
Speed of the train, u = 90 km/h.
Final speed, v = 0.
Step 2: Finding the distance traveled by train:
Speed of the train, $u = 90 k m / h = \frac{90 \times 1000}{3600} = 25 m / s .$
The train will stop after some time, so its final velocity will be zero.
From the newton's third equation of motion;
$v^{2} - u^{2} = 2 a s .$ ……(1)
Where v and u are the final and initial speeds respectively.
‘a’ is acceleration and 's' is the distance traveled.
From equation (1)
$0 - u^{2} = 2 a s - u^{2} = 2 a s s = \frac{u^{2}}{2 a} = \frac{- (25^{2})}{2 \times (- 0.5)} = 625 m .$
The train will stop after 625 m.

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A train is traveling at 90km/h.The brakes are applied so as to produce a uniform acceleration of -0.5m/s2.Find how far the train goes before it stops.

A train is traveling at $90 k m / h .$ The brakes are applied so as to produce a uniform acceleration of $- 0.5 m / s^{2} .$ Find how far the train goes before it stops.