Question

A bus travels for 2 hr at a constant speed of 40 km hr" and then returns to the starting point at a constant speed of 60 km hr"! Average speed of the bus is km hr!​

Answer 1

Average Speed:

• It is defined as the ratio of total distance travelled to the total time taken.

• Distance of 1st part = 40 × 2 = 80 km.

• Since bus returns back to same position, total distance is 80+80 = 160 km.

Let Average Speed be v :

$\sf \: v = \dfrac{total \: distance}{total \: time}$

$\sf \implies\: v = \dfrac{160}{ t_{1} + t_{2}}$

$\sf \implies\: v = \dfrac{160}{ 2+ \dfrac{80}{60} }$

$\sf \implies\: v = \dfrac{160}{ 2+ 1.33}$

$\sf \implies\: v = \dfrac{160}{ 3.33}$

$\sf \implies\: v = 48.04 \: kmph$

So, average speed is 48.04 km/hr.

Answer 2

Given:

Speed of bus while travelling from one point to other ($v_{1}$) = $40 Km/hr$

Speed of bus after returning to same point ($v_{2}$) = $60 Km/hr$

To Find:

Average (av.) speed of bus =?

Solution:

Speed is distance covered while travelling from one point to another.

Average-speed of the bus-

Average = $\dfrac{2v_{1}v_{2} }{v_{1}+v_{2} }$

 $v_{1}$ = 40 ,  $v_{2}$  = 60

                = 2 × 60 × 40 / 60+40

               = 4800/100

               = 48 $Km/hr$

Average-speed of the bus is 48 $Km/hr$