## Related questions with answers

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s². How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of 1.65 m/s². How long does it take to come to a stop from its top speed? (c) In emergencies, the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency acceleration in meters per second squared?

Solution

Verified(a) Train starts from rest. So, $v_0 = 0\:m.s^{-1}$

Its top speed is $v=80.0\:\dfrac{km}{h} = 80.0 \times \dfrac{10^3\:m}{3600\:s} = 22.22\:m.s^{-1}$

where $1\:km = 10^3\:m$ and $1\:h = 3600\:s$

Acceleration of train is $a=1.35\:m.s^{-2}$

We have the following kinematic equation,

$v = v_0 + at$

Substituting the given values

$22.22 = 0 + 1.35 t$

$t = \dfrac{22.22}{1.35} \:\approx\:16.5\:s$

So the time taken to reach the top speed from rest at the given acceleration is $16.5\:s$.

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