Question

A car travels a certain distance with a speed of 30km/h and returns with a speed of 45km/h. Calculate the average speed

Answer 1

GIVEN :–

• A car travels a certain distance with a speed of 30km/h and returns with a speed of 45km/h.

TO FIND :–

• Average speed = ?

SOLUTION :–

• Let a distance x km be travelled with a speed of 30 km/hr.

• Now let's find taken time –

$\\ \implies \bf Time=\dfrac{Distance}{Speed}$

$\\ \implies \bf T_1=\dfrac{x}{30}hr$

• Similarly distance x km be travelled with a speed of 45 km/hr.

• Time taken –

$\\ \implies \bf T_2=\dfrac{x}{45}hr$

• Total taken time –

$\\ \implies \bf T = T_1 + T_2$

$\\ \implies \bf T =\dfrac{x}{30}+ \dfrac{x}{45}$

• Now let's find Average speed –

$\\ \implies \bf Average \: \: speed =\dfrac{(x + x)}{ \left(\dfrac{x}{30}+ \dfrac{x}{45} \right)}$

$\\ \implies \bf Average \: \: speed =\dfrac{2x}{ \left(\dfrac{45x + 30x}{30 \times 45} \right)}$

$\\ \implies \bf Average \: \: speed =\dfrac{2x}{ \left(\dfrac{75x}{30 \times 45} \right)}$

$\\ \implies \bf Average \: \: speed =\dfrac{2x}{ \left(\dfrac{75x}{1350} \right)}$

$\\ \implies \bf Average \: \: speed =\dfrac{(2x)(1350)}{75x}$

$\\ \implies \bf Average \: \: speed =\dfrac{2700}{75}$

$\\ \implies \red{\large{\boxed{\bf Average \: \: speed =36 \: Km/hr}}}$

Answer 2

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ A car travels a certain distance with a speed of 30km/h and returns with a speed of 45km/h

${\large{\bold{\rm{\underline{To \; find}}}}}$

★ The average speed

${\large{\bold{\rm{\underline{Solution}}}}}$

★ The average speed = 36 km/h

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

_________________

~ According to the question let us take a as assumption to travelled distance with a speed of 30 km/h

~ Firstly let us find the taken time..!

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

$\begin{gathered}\\ \longrightarrow \sf Time \: = \dfrac{a}{30} hr. \\\end{gathered}$

_________________

~ According to the question let us take a as assumption to travelled distance with a speed of 45 km/h

~ Firstly let us find the taken time..!

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

$\begin{gathered}\\ \longrightarrow \sf Time \: = \dfrac{a}{45} hr. \\\end{gathered}$

_________________

~ Henceforth, total time taken is,

$\begin{gathered}\\ \longrightarrow \sf Time \: = \dfrac{a}{30} + \dfrac{a}{45} \\\end{gathered}$

~ Now let's find the average speed..!

$\begin{gathered}\\ \longrightarrow \sf Average \: speed = \dfrac{(a + a)}{\left(\dfrac{a}{30}+ \dfrac{a}{45} \right)}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: speed = \dfrac{2a}{\left(\dfrac{45a + 30a}{30 \times 45} \right)}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: speed = \dfrac{2a}{\left(\dfrac{75a}{30 \times 45} \right)}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: \: speed =\dfrac{2a}{ \left(\dfrac{75a}{1350} \right)}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: speed =\dfrac{(2a)(1350)}{75a}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: speed =\dfrac{2700}{75}\\\end{gathered}$

$\begin{gathered}\\ \longrightarrow \sf Average \: speed =\: 36 \: km/hr\\\end{gathered}$

${\large{\bold{\rm{\underline{Additional \; information}}}}}$

Distance = It is the length of actual path covered by a moving object in a given time interval.

Displacement = Shortest distance covered by a body in a definite direction is called displacement.

→ Displacement may be positive, negative or zero whereas distance is always positive.

→ Distance is a scaler quantity whereas displacement is a vector quantity both having the same unit.

→ In general magnitude of displacement ≤ Distance.

Speed = Distance travelled by a moving object in unit time interval is called speed i.e., speed Distance/Time. It's scaler quantity and it's SI unit is metre/second