A student starts from his physics classroom considered to be origin, walks $20 m$ down the corridor, then stops, turns around and walks $5 m$ back to his classroom. He stops $15 m$ from the door. Total time of motion is $25.0 s e c$ . What is his average speed?

1 m / s

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2 m / s

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3 m / s

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4 m / s

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Solution

The correct option is A $1 m / s$

Path length until he turns back = $20 m$
Path length after he turns back = $5 m$
So, total distance travelled = $25 m$

$Average speed = \frac{total distance travelled}{time taken}$

$= \frac{20.0 + 5.0}{25.0} = 1.0 m / s$

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Similar questions

A boy walks from his classroom to the bookshop along a straight corridor towards North. He covers a distance of 20 m is 25 seconds to reach the bookshop. After buying a book, he travels the same distance at the same time to reach back in the classroom. Find (a) average speed, and (b) average velocity, of the boy.

Q. A student starts from his physics classroom considered to be origin, walks

20 m

down the corridor, then stops, turns around and walks

5 m

back to his classroom. He stops

15 m

from the door. Total time of motion is

25.0 s e c

. What is his average speed?

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A student starts from his physics classroom considered to be origin, walks 20 m down the corridor, then stops, turns around and walks 5 m back to his classroom. He stops 15 m from the door. Total time of motion is 25.0 sec. What is his average speed?

The correct option is A 1 m/s Path length until he turns back = 20 m Path length after he turns back = 5 m So, total distance travelled = 25 m Average speed=total distance travelledtime taken =20.0+5.025.0=1.0 m/s

A student starts from his physics classroom considered to be origin, walks $20 m$ down the corridor, then stops, turns around and walks $5 m$ back to his classroom. He stops $15 m$ from the door. Total time of motion is $25.0 s e c$ . What is his average speed?

The correct option is A $1 m / s$

Path length until he turns back = $20 m$
Path length after he turns back = $5 m$
So, total distance travelled = $25 m$

$Average speed = \frac{total distance travelled}{time taken}$

$= \frac{20.0 + 5.0}{25.0} = 1.0 m / s$