Question

a person walks along a straight road from his house to a market 2.5 kilometres away with a speed of 5 kilometres per hour and instantly turns back and reaches his house with a speed of 7.5 km per hour the average speed of the person during the time interval 0 to 50 minutes is(in m/s)​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Distance = 2.5 km.

Speed (u)= 5 km/hr.

(He goes to the market)

Spedd(v) = 7.5 km/hr.

(He comes back from market)

$\huge{\bold{\underline{Explanation:-}}}$

$\large{t_1 = \frac{Distance}{Speed}}$

$\large{ \to t_1 = \frac{2.5}{5}}$

$\large{ \to t_1 = 0.5 hour = 30\: minutes}$

$\large{t_2 = \frac{Distance}{Speed}}$

$\large{ \to t_2 = \frac{2.5}{7.5}}$

$\large{ \to t_2 = 0.3hour = 20\: minutes}$

So, he travels for 50 min and for that we have to find average speed.

so,by average speed formula.

$\huge{\boxed{\boxed{Average\: Speed = \frac{Total\: Distance}{Total\: Time}}}}$

$\large{V_{av} = \frac{ 5 + 5}{0.5 + 0.3}}$

$\large{ \to V_{av} = \frac{10}{0.8}}$

$\large{ \to V_{av} = \frac{100}{8}}$

$\large{ \to V_{av} = \frac{25}{4} Kmh^{-1}}$

Converting it to m/s.

$\large{V_{av} = \frac{25}{4} \times \frac{5}{18}}$

$\large{ \to V_{av} = \frac{125}{72}\: ms^{-1}}$

$\large{ \to V_{av} = 1.73 ms^{-1}}$

$\huge{\boxed{\boxed{V_{av} = 1.73 \: ms^{-1}}}}$

So,the average speed is 1.73 m/s.