A nice physics based question:

6 viewed last edited 5 years ago

A biker accelerates from rest at a constant 3m/s2, for sometime, continues with uniform velocity for the same time and retards for some more time at 6m/s2. If the bike has travelled for 25 minuets, what is the total distance covered by him?  

Vivekanand Vellanki

Plotting the biker's motion in a graph:

The portion from O to A shows uniform acceleration. The portion from A to B shows uniform velocity; and the portion from B to C shows uniform retardation.

In this figure, we know the following:

t1 + t2 + t3 = 25

t1 = t2

Also, t3 = t1/2 because the deceleration is twice that of the acceleration.

This gives t1 = t2 = 10 min; and t3 = 5 mins

v = 0 + 3*10*60 = 1800 m/s.

The distance covered is the area under the curve.

Distance = \frac{1}{2}\cdot v\cdot t1+v\cdot t2+\frac{1}{2}\cdot v\cdot t3=v\left(\frac{t1}{2}+t2+\frac{t3}{2}\right)