Question

A man covers a distance from a point P to a point Q in a straight line at a speed of 4m/s and returns back along the same path at a speed of 8m/s .What is his average speed? What is his average velocity?

Answer 1

In the above Question , the following information is given -

A man covers a distance from a point P to a point Q in a straight line at a speed of 4m/s and returns back along the same path at a speed of 8m/s .

To find -

1. What is his average speed ?

2. What is his average velocity ?

Solution -

Here , let P and Q be two arbitary points and let the distance between them be x metres.

Refer to the attachment alongside.

Now ,

Initially, the man goes at a speed of 4 m / s.

We have earlier assumed that the path is x m long.

Speed Of man - 4 m/ s.

The man takes ( ¼ ) seconds to traverse 1 m .

So , to traverse x m , he will take , ( ¼ ) × x seconds or ( x / 4 ) seconds.

Now , he returns at a speed of 8 m / s.

Speed Of man - 8 m/ s.

The man takes ( ⅛ ) seconds to traverse 1 m .

So , to traverse x m , he will take , ( ⅛ ) × x seconds or ( x / 8 ) seconds.

Now,

Distance travelled by the man -

=> ( x + x ) m

=> ( 2x ) metres.

The displacement of the man is 0 as he returns to his initial point .

So ,

Average Speed

=> [ Total distance travelled ] / [ Total time taken ]

=> [ 2 x ] / [ ¼ x + ⅛ x ]

=> [ 2 x ] / [ ⅜ x ]

=> ( 16 / 3 ) m / s.

Average velocity -

=> [ Total displacement ] / [ Total time taken ]

=> 0 m / s .

This is the required answer .

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Answer 2

Explanation:

To find:

★ average speed

★average velocity

Solution:

$t_1 = \frac{x}{4}$

$t_2 = \frac{x}{8}$

$x = 4t_1$

$x = 8t_2$

$4t_1 = 8t_2$

$\frac{t_1}{t_2} = \frac{8}{4}$

$t_1 = 2t_2$

$v_1 = 4ms ^{ - 1}$

$v_2 = 8ms ^{ - 1}$

$t_1 = 2t$

$t_2 = t$

Average speed=total distance/total time

$\implies \: \frac{(v_1 \times \: t_1 ) + v_2 \times t_2}{t_1 +t_2 }$

$= \frac{4 \times 2t \times 8 \times t}{2t + t}$

$= \frac{16}{3} ms ^{ - 1}$

Average velocity=total displacement/total time

=0 ms-¹.