An athlete runs 300 metres up a hill at a steady speed of 3 m/s. She then immediately runs the same distance down the hill at a steady state of 6 m/s. What is her average speed for the 600-metre run?
(A) 2 m/s
(B) 3 m/s
(C) 4 m/s
(D) 4.5 m/s
Answer
641.7k+ views
Hint:Here the person goes and covers some distance at a constant speed. After reaching the point he turns back and covers the same distance but at some another speed which is also constant. We need to find the average speed during the whole journey
Complete step by step answer:
The first leg of the journey:
Distance covered= 300 m
Speed = 3 m/s
Using the formula
\[\begin{align}
& speed=\dfrac{d}{t} \\
& t=\dfrac{d}{v} \\
\end{align}\]
\[t=\dfrac{300}{3}=100s\]
So the time needed to cover the first leg of the journey is 100s, now we come to the second leg of the journey.
Distance covered= 300 m
Speed = 6 m/s
Using the formula
\[\begin{align}
& speed=\dfrac{d}{t} \\
& t=\dfrac{300}{6}=50s \\
\end{align}\]
So the time needed to cover the first leg of the journey is 50 s, now looking at the whole journey together.
Total distance covered= 300+300= 600 m
Total time taken, t= (100+50)= 150 s
So average speed of the journey= \[\dfrac{Distance}{Time}=\dfrac{600}{150}=4m/s\]
The average speed comes out to be 4 m/s. So, the correct option is C
Note: If in this problem we would have been given to find out the average velocity, then our answer would have been zero because the net displacement, in this case, is zero because the initial and final positions are the same. Displacement is the shortest path between the initial and final position.
Complete step by step answer:
The first leg of the journey:
Distance covered= 300 m
Speed = 3 m/s
Using the formula
\[\begin{align}
& speed=\dfrac{d}{t} \\
& t=\dfrac{d}{v} \\
\end{align}\]
\[t=\dfrac{300}{3}=100s\]
So the time needed to cover the first leg of the journey is 100s, now we come to the second leg of the journey.
Distance covered= 300 m
Speed = 6 m/s
Using the formula
\[\begin{align}
& speed=\dfrac{d}{t} \\
& t=\dfrac{300}{6}=50s \\
\end{align}\]
So the time needed to cover the first leg of the journey is 50 s, now looking at the whole journey together.
Total distance covered= 300+300= 600 m
Total time taken, t= (100+50)= 150 s
So average speed of the journey= \[\dfrac{Distance}{Time}=\dfrac{600}{150}=4m/s\]
The average speed comes out to be 4 m/s. So, the correct option is C
Note: If in this problem we would have been given to find out the average velocity, then our answer would have been zero because the net displacement, in this case, is zero because the initial and final positions are the same. Displacement is the shortest path between the initial and final position.
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