A jogger moves 500 m in 2 minutes and next 1000 m in 30 s on the same straight path. What is his average speed?
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A jogger moves 500 m in 2 minutes and next 1000 m in 30 s on the same straight path. What is his average speed?
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Given :-
A jogger moves 500 meters in 2 minutes and next 1000 meters in 30 seconds .
The direction of motion is straight
Required to find :-
Formula used :-
[tex] \boxed{ \sf{ \red{average \: speed = } \green{ \frac{total \: distance}{total \: time \: taken} }}} \bigstar[/tex]
Solution :-
Given information :-
A jogger moves 500 meters in 2 minutes and next 1000 meters in 30 seconds .
we need to find the average speed .
From the given data we can conclude that ;
Case - 1
Distance ( d1 ) = 500 meters
Time ( t1 ) = 2 minutes ( 120 seconds )
Case - 2
Distance ( d2 ) = 1000 meters
Time ( t2 ) = 30 seconds
Since,
The S.I. unit of speed is m/s
Let's time ( t1 ) from minutes 10 seconds
So,
2 minutes = ? seconds
=> 2 x 60
=> 120
2 minutes = 120 seconds
Using the formula ;
[tex] \: \boxed{ \sf{ \red{average \: speed = } \green{ \frac{total \: distance}{total \: time \: taken} }}} \bigstar[/tex]
Here,
Total distance = d1 + d2
=> 500 meters + 1000 meters
=> 1500 meters
Similarly,
Total time taken = t1 + t2
=> 120 seconds + 30 seconds
=> 150 seconds
Substituting the values ;
[tex] \rightarrowtail \rm average \: speed = \dfrac{1500 \: meters}{150 \: seconds} \\ \\ \rightarrowtail \rm average \: speed = 10 \: m {s}^{ - 1} [/tex]
Therefore,
Average speed of the jogger is 10 m/s
Additional information :-
Speed is a scalar quantity .
Velocity is a vector quantity .
Speed requires only magnitude but not direction .
Velocity requires both magnitude and direction .
The scenario mentioned above in the question tells about the velocity because it is travelling in a straight path .
This states the fact that ;
In some cases we can take velocity in terms of speed but we can't take speed in the terms of velocity