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A boy walks on a straight road from his house to the market 2.5 km with a speed of $5\; km{{h}^{-1}}$ . Finding the market closed he instantly turns and walk back with a speed of $7.5\; km{{h}^{-1}}$ .What is the average speed and average velocity of the boy between $t=0 \; to \; t=50\text{ }min$ ?

Option: 1
$0,0$

Option: 2
$6\: km{{h}^{-1}},0$

Option: 3
$0,6km{{h}^{-1}}$

Option: 4
$6km{{h}^{-1}},6km{{h}^{-1}}$

Answers (1)

best_answer

Time taken by the boy to go from his home to the market,

${{t}_{1}}=\frac{2.5km}{5km{{h}^{-1}}}=\frac{1}{2}h$

Time taken by the boy to return back from the market to his home,

${{t}_{2}}=\frac{2.5km}{7.5km{{h}^{-1}}}=\frac{1}{3}h$

$\therefore \text{Total time taken}={{t}_{1}}+{{t}_{2}}=\frac{1}{2}h+\frac{1}{3}h=\frac{5}{6}h=50\min$

In t=0 to t=50 min

$\text{Total distance travelled}=2.5km+2.5km=5km$

$\text{Displacement}=0$ (As the boy return back home)

$\mathrm{Avaerage \: Speed=\frac{\text{Total distance travelled}}{\text{Total time taken}}}$

$\vec{v}=\frac{5km}{\frac{5}{6}h}=6km{{h}^{-1}}$

$\text{Average velocity}=\frac{\text{Displacement}}{\text{Time taken}}=0$

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shivangi.bhatnagar

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