Question

A car travels with the speed of 40km/h from a city A to the city B and returns back with
a speed 60km/h. Calculate the average speed of car (1) average velocity of car.
16 Frequency of transverse vibrations in a stretched string depends on the length!​

Answer 1

~$\small \blue {\sf{Answer }}$✨❤

Let the each half of the running time be t.

During first half time, speed of car is 80 km/h

Distance covered d 1 =80t km

During second half time, speed of car is 40 km/h

Distance covered d 2

=40t km

Total distance covered d=d

$d=d1+d2=80t+40t=120t</p><p>$

Total time taken T=2t

Option C is correct.

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Answer 2

Explanation:

Speed of car = 40km/h from city A to B

Speed of car = 60km/h from city B to A

Let the distance between A to B is 'AB' km

and time taken from A to B is T1

then

$T _{1} = \frac{(distnce \: between \: a \: to \: b)}{(speed \: of \: car \: from \: a \: to \: b)} \\ \\ T _{1} = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>4</strong><strong>0</strong><strong>}$

similarly time taken from B to A T2

then

$T _{2} = \frac{(distnce \: between \: a \: to \: b)}{(speed \: of \: car \: from \: a \: to \: b)} \\ \\ T _{2} = \frac{</strong><strong>AB</strong><strong>}{60}$

Hence average speed =

$\frac{(total \: trvelled \: distance)}{total \: taken \: time} \\ \frac{</strong><strong>AB</strong><strong> + </strong><strong>AB</strong><strong>}{ \frac{</strong><strong>AB</strong><strong>}{40} + \frac{</strong><strong>AB</strong><strong>}{60} } \\ \frac{</strong><strong>2AB</strong><strong> }{ \frac{</strong><strong>5AB</strong><strong>}{120} } \\ \frac{</strong><strong>2AB</strong><strong> \times 120}{</strong><strong>5AB</strong><strong>} \\ 48 \: km</strong><strong>/</strong><strong>h$

2.. Average velocity become zero because car reached to point A

So diaplacement be zero..

hence velocity zero.