Problem 11 on Motion in a Straight Line

Description:

Problem

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes.

A man cycling with a speed of 20 km/h in the direction A to B notices that a bus goes past him every 18 minutes in the direction of his motion, and every 6 minutes in the opposite direction.

What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

Solution

Assume the cyclist to be moving East.

Therefore,

• buses moving East approach cyclist from behind.
• buses moving West approach cyclist from front.

Following sign convention is assumed for this problem −

Assume that speed (magnitude of velocity) of buses with respect to ground is (v) km/h.

Velocity of buses moving East is assumed to be v→_E.

Velocity of buses moving West is assumed to be, v→_W.

Velocity of cyclist is assumed to be, v→_C.

Therefore,

Velocity of buses moving East w.r.t to cyclist,

v→_EC = (v→_E - v→_C) = (v - 20) km/h

Velocity of buses moving West w.r.t to cyclist,

v→_WC = (v→_W - v→_C) = (-v - 20) = -(v + 20) km/h

Now,

Because buses leave in either direction every ‘T’ minutes, separation between any two busses in the same direction is,

Separation = (d) km = vT

Buses moving East pass the cyclist every 18 minutes. Therefore, with respect to cyclist,

d/(v - 20) = 18 minutes

Buses moving West pass the cyclist every 6 minutes. Therefore, with respect to cyclist,

-d/-(v + 20) = 6 minutes

Note: Busses travelling West will have negative displacement due to chosen sign convention. Hence, the negative sign in the numerator of second equation.

Taking their ratio,

(v + 20)/(v - 20) = 3

v = 40 km/h

Now,

To calculate ‘𝑇’, substituting the value of ‘𝑑’ as ‘𝑣𝑇’ in the first equation, d/(v - 20) = 18 minutes,

vT/(v - 20) = 18 minutes

Putting the value of ‘v’,

40 × T/(40 - 20) = 18 minutes

2 × T = 18 minutes

T = 9 minutes