Question

A girl walks to her school ar a distance of 1
km with a speed of 2 kmp) and comes back
with a speed of 3 kmph. The average speed

Answer 1

Given: A girl walks to her school ar a distance of 1 km with a speed of 2 kmph & comes back with a speed of 3 kmph.

Need to find: Average speed of the girl.

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¤ To calculate average speed we need find out total distance and total time, So let's find them.

$\dag \: {\red{\pmb{\frak{Total \: distance : }}}}$

Let point X be girls home and let point Y be girl's school.

$\begin{gathered} : \implies\sf D_{(Total\:distance)}= XY + YX \\\\\\ : \implies \sf D_{(Total\:distance)} = 1 km + 1 km \\\\\\ : \implies \sf D = 2 \ km \end{gathered}$

∴ Total distance = 2 km.

$\dag \: {\red{\pmb{\frak{Total \: time : }}}}$

$\begin{gathered} \dag \: {\pmb{ \sf{ Time= \dfrac{Distance}{Speed}}}} \end{gathered}$

$\begin{gathered} : \implies\sf T_{1}= \dfrac{1}{2}\\\\\\: \implies \pmb{\sf 1.5 hr }\\\\\\: \implies\sf T_{2}= \dfrac{1}{3}\\\\\\: \implies\pmb{ \sf{0.33 hr}}\end{gathered}$

∴ Total time is 1.5 + 1.33 i.e 0.83 hr.

⠀

$\dag \: {\red{\pmb{\frak{Average\: speed : }}}}$

$\begin{gathered} \dag \: \: {\pmb{ \sf{ V avg= \dfrac{Distance}{Time}}}} \end{gathered}$

$\begin{gathered} : \implies\sf V avg = \dfrac{1}{0.83} \\\\\\ : \implies \sf V avg = \dfrac{2}{0.83} \\\\\\ : \implies \pmb{\sf \red{ V avg = 2.41 km^{-1}}} \end{gathered}$

$\therefore \sf Average \: speed \: of \: the \: girl \: is \: \pmb{\sf{2.41 km^{-1} }}, Respectively.$

Answer 2

Explanation:

hopefully this helps u champ, rock in ur studies