Question

A man walks 5km and 10 km at a speed of 10m/s and 5m/s respectively. His average speed is,  (1 Point) () 1  −1(a) 1 m s−1 () 25  −1(b) 25 ms −1 () 54  −1(c) 54  ms−1​

Answer 1

$\Large {\underline { \bf {Answer :}}}$

⠀⠀⠀⠀⠀⠀» Average speed = 6 m/s «

$\Large {\underline { \bf {Explanation :}}}$

As per the provided information in the given question, we have:

• A man walks 5 km and 10 km at a speed of 10m/s and 5m/s respectively.

We've been asked to calculate average speed.

In order to calculate the average speed, firstly we need to calculate total distance and total time taken.

$\implies{\sf{Total \; Distance = (5 + 10) \;km}}$

$\implies{\boxed{\sf{Total \; Distance = 15 \;km}}}$

Now, we have to calculate total time taken. Total time taken will be the sum of the time taken to cover first 5 km and time taken to cover next 10 km. So,

• S₁ = 5 km
• V₁ = 10 m/s
• S₂ = 10 km
• V₂ = 5 m/s

So,

$\implies{\sf{Total \; Time = T_1 + T_2}}$

• Time = Distance ÷ Speed

$\implies{\sf{Total \; Time = \dfrac{S_1}{V_1} + \dfrac{S_2}{V_2} }}$

$\implies{\sf{Total \; Time = \Bigg \{ \dfrac{5}{10} + \dfrac{10}{5} \Bigg \} \; s }}$

$\implies{\sf{Total \; Time = \Bigg \{ \dfrac{1}{2} + 2\Bigg \} \; s }}$

$\implies{\sf{Total \; Time = \Bigg \{ \dfrac{1+ 4}{2} \Bigg \} \; s }}$

$\implies{\boxed{\sf{Total \; Time = \dfrac{5}{2} \; s }}}$

Now, as we know that :

$\bigstar{\underline{\boxed{ \pmb{\frak{Speed}}_{\pmb{\frak{(Avg)}}} = \dfrac{\pmb{\frak{Total \; Distance}}}{\pmb{\frak{Total \; Time}}} }}}$

$\implies{\sf{Speed_{(Avg)} = \Bigg \{ 15 \div \dfrac{5}{2} \Bigg \} \; m \: s^{-1} }}$

$\implies{\sf{Speed_{(Avg)} = \Bigg \{ 15 \times \dfrac{2}{5} \Bigg \} \; m \: s^{-1} }}$

$\implies{\sf{Speed_{(Avg)} = \Bigg \{ 3 \times 2 \Bigg \} \; m \: s^{-1} }}$

$\implies{\underline{\bf{Speed_{(Avg)} = 6 \; m \: s^{-1} }}} \; \bigstar$

Therefore, the average speed of the man is 6 m/s.