Question

Joseph jogs from one end A to the other end B of a straight

300 m

road in

2 m i n

30 s

and then turns around and jogs

100 m

back to point C in another

1 m i n

. What are Joseph’s average speeds and velocities in jogging from A to C?

Accepted Answer

From end A to end C

$Average speed = \frac{Total distance covered}{Total time taken}$

Total distance covered by Joseph from A to C = Distance covered from A to B + Distance covered from B to C.

$Total distance = 300 + 100 = 400 m$ .

Total time taken by Joseph = Time taken by Joseph to travel from A to B + Time taken by Joseph to travel from B to C.

$Total time = 150 + 60 = 210 s$ .

$Average speed = 400 / 210 = 1 .90 m / s$

We know that, $Avera g e velocity = \frac{Displacement}{time interval}$

Displacement is the shortest distance between two points and as the particle is coming back to point C, so displacement is equal to AC.
Displacement $AC = A B - B C = 300 - 100 = 200 m$ .

Time interval = Time taken to travel from A to B + Time taken to travel from B to C.
$Time interval = 150 + 60 = 210 s$ .

$Average velocity = 200 / 210 = 0 .95 m / s$ .

The average speed of Joseph from A to C is $1.90 m / s,$ and his average velocity is $0.95 m / s .$

NCERT Solutions for Chapter: Motion, Exercise 6: Exercises

NCERT Science Solutions for Exercise - NCERT Solutions for Chapter: Motion, Exercise 6: Exercises

Questions from NCERT Solutions for Chapter: Motion, Exercise 6: Exercises with Hints & Solutions

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