Question

A car travels from A to B at a speed of 40 km h-1 and returns back on the same track at a speed of 60 km h-1 Average velocity will be [NCERT Pg. 42] (1) 60 km h-1 (2) Zero (3) 48 km h-1 (4) 50 km h-1 ..​

Answer 1

Answer:

Explanation:

Let the distance between A and B be x.

The time taken by the car to travel from A to B= x/40.

( Since speed= distance/time. so time = distance/speed)

The time taken by the car to travel from B to A=x/30

Average speed = total distance/ total

time

= 2x/70x/120

= 24/7 Km/hr

The displacement of the car is zero because its final and initial point is same and hence the average velocity is also zero

Answer 2

Solution:

Average velocity = zero.

Using concepts:

• Total distance is given by,

• ${\small{\underline{\boxed{\sf{Total \: distance \: = 1st \: distance \: + 2nd \: distance}}}}}$

• Average velocity is given by,

• ${\small{\underline{\boxed{\sf{a_{av} \: = \dfrac{s}{t}}}}}}$

Where, a_{av} denotes average velocity, s denotes displacement and t denotes total time taken.

• Time is given by,

• ${\small{\underline{\boxed{\sf{t \: = \dfrac{s}{v}}}}}}$

Where, t denotes time taken, s denotes distance and v denotes speed.

• Displacement:

• Displacement is the shortest distance between the initial and final position.

• If initial and final positions are same then displacement must be zero.

Assumption:

• Let the distance travelled as a

Required solution:

~ Firstly finding total distance!

→ As it travel and return too and take same distance. Therefore,

• Total distance = 2a.

~ Now let us find out the time taken!

$\begin{gathered}:\implies \sf Time \: taken \: = \dfrac{Distance}{Speed} \\ \\ :\implies \sf t \: = \dfrac{s}{v} \\ \\ :\implies \sf t \: = \dfrac{a}{40} + \dfrac{a}{60} \\ \\ \leadsto \sf Taking \: LCM \: of \: 40 \: and \: 60 \\ \\ :\implies \sf t \: = \dfrac{3 \times a + 2 \times a}{120} \\ \\ :\implies \sf t \: = \dfrac{3a + 2a}{120} \\ \\ :\implies \sf t \: = \dfrac{5a}{120} \\ \\ :\implies \sf Time \: = \dfrac{5a}{120} \: hour\end{gathered}$

Therefore, time is 5a/120 hr.

~ Now let us calculate the displacement.

We know that if initial and final positions are same then displacement must be zero.

And here according to the question initial position and final position are same. Therefore, displacement = 0.

~ Now let us calculate the average velocity by using suitable formula!

$:\implies \sf a_{av} \: = \dfrac{s}{t} \\ \\ :\implies \sf a_{av} \: = \dfrac{0}{5a/120} \\ \\ :\implies \sf a_{av} \: = 0 \: kmph \\ \\ :\implies \sf Average \: velocity \: = 0 \: kmph$