Question

A car traveling A certain distance with a speed of 50 kilometre per hour and returns with a speed of 40 kilometre per hour. calculate the average speed of the whole journey​

Answer 1

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Average Speed has been used. Average speed is the Total Distance covered divided by Total Time Taken in the Journey. Here we will take the Distance covered to be a variable quantity x and speed as s. Then, we can apply the formula and find the answer.

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★ Formula Used :-

$\: \\ \large{\boxed{\boxed{\sf{Time \: = \: \bf{\dfrac{Distance}{Speed}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Time \: taken \: while \: going, \: (T_{1}) \: = \: \bf{\dfrac{x \: \: Km}{50 \: Kmhr^{-1}}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Time \: taken \: while \: returning, \: (T_{2}) \: = \: \bf{\dfrac{x \: \: Km}{40 \: Kmhr^{-1}}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{Average \: Speed, \: (s_{(av)}) \: = \: \bf{\dfrac{Total \: Distance}{Total \: Time \: Taken}}}}}}$

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★ Question :-

A car traveling A certain distance with a speed of 50 kilometre per hour and returns with a speed of 40 kilometre per hour. Calculate the average speed of the whole journey.

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★ Solution :-

Given,

» Speed of car while going = s = 50 Km/hr

» Speed of car while returning = s' = 40 Km/hr

We know that distance while going will be equal to distance while coming.

» Distance while going = x Km = Distance while returning

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~ For time taken while going :-

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Time \: = \: \bf{\dfrac{Distance \: (x)}{Speed \: (s)}}}}$

$\: \\ \qquad \large{\sf{:\longrightarrow \: \: \: Time \: taken \: while \: going, \: (T_{1}) \: = \: \bf{\dfrac{x \: \cancel{Km}}{50 \: \cancel{Km}hr^{-1}}}}}$

$\: \\ \: \large{\boxed{\boxed{\sf{Time \: taken \: while \: going, \: T_{1} \: = \: \bf{\dfrac{x}{50} \: \: hr}}}}}$

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~ For time taken while returning :-

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: Time \: = \: \bf{\dfrac{Distance \: (x)}{Speed \: (s')}}}}$

$\: \\\large{\sf{Time \: taken \: while \: returning, \: (T_{2}) \: = \: \bf{\dfrac{x \: \cancel{Km}}{40 \: \cancel{Km}hr^{-1}}}}}$

$\:\: \\ \large{\boxed{\boxed{\sf{Time \: taken \: while \: returning, \: T_{2} \: = \: \bf{\dfrac{x}{40} \: \: hr}}}}}$

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~ For Average Speed of the journey :-

• Total distance covered = Distance while going + Distance while returning

$\: \\ \large{\sf{\rightarrow \: \: Total \: Distance \: covered \: = \: \tt{x \:\; Km \;\: + \;\: x \:\; Km } \: \: = \: \: \underline{\underline{\bf{2x \:\; Km}}}}}$

• Total Time Taken = Time taken while going + Time taken while returning

$\: \\ \large{\sf{\rightarrow \: \: Total \: Distance \: covered \: = \: \tt{\:\; T_{1} \:\;+\:\; T_{2} \:\; \; = \:\;\: \dfrac{x}{50} \:\; hr \;\: + \;\: \dfrac{x}{40} \:\; hr} \: \: = \: \: \bf{\dfrac{9x}{200} \: \: hr}}}$

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Now average speed is given as :-

$\: \\ \qquad \large{\sf{:\longrightarrow \: \: \: Average \: Speed, \: (s_{(av)}) \: = \: \bf{\dfrac{Total \: Distance}{Total \: Time \: Taken}}}}$

$\: \\ \qquad \large{\sf{:\longrightarrow \: \: \: Average \: Speed, \: (s_{(av)}) \: = \: \bf{\dfrac{2\cancel{x} \: Km}{\dfrac{9\cancel{x}}{200} \: hr}}} \: \: = \: \: \bf{\dfrac{400}{9} \: \: Kmhr^{-1}}}$

$\: \: \\ \large{\underline{\underline{\rm{Thus, \: average \: speed \: of \: journey \: is \: \: \boxed{\bf{\dfrac{400}{9} \: \: \: Kmhr^{-1}} \;\;\tt{or} \:\;\;\;\bf{44.44\: \: \: Kmhr^{-1}}}}}}}$

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$\: \: \qquad \large{\underbrace{\underbrace{\tt{More \;\; to \;\; know \;\;\; :-}}}}$

• Speed is the distance covered in a certain time in any direction. Its a scalar quantity.

• Velocity is the displacement covered in per unit time in a certain direction. Its a vector quantity.

• Acceleration is the rate of change of velocity in a unit time.

• Displacement is the shortest path to a end point. Its a vector quantity.

• Distance is the length of the path from a point to a end point. Its a scalar quantity.

Answer 2

$\huge\bold{\underbrace{Answer!!!}}$

Given :

• A car travelling a certain distance with a speed of 50 km/h.
• Again car returns with a speed of 40 km/h.

To find :

• Average speed of the whole journey.

Formulae used :

• Time = $\bf\dfrac{Distance}{Speed}$
• Average speed = $\bf\dfrac{Total \ distance \ travelled}{Total \ time \ taken}$

Solution :

Firstly to find out the average speed, we need Total distance travelled by the car and Total time taken by the car.

So we have to find out the values of distance and time.

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Let the distance be 'x'

${\implies}$Time taken by the car at a speed of 50 km/h = $\bf\dfrac{Distance}{Speed}$

${\implies}$ $\bf\dfrac{x}{50}$ h

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${\implies}$Time taken by the car at a speed of 40 km/h = $\bf\dfrac{Distance}{Speed}$

${\implies}$ $\bf\dfrac{x}{40}$ h

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${\implies}$Total time = $\bf\dfrac{x}{50}$ h + $\bf\dfrac{x}{40}$ h

= $\bf\dfrac{4x + 5x}{200}$ h

= $\bf\dfrac{9x}{200}$ h

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${\implies}$Total distance = x + x = 2x km

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${\implies}$Average Speed = $\bf\dfrac{Total \ distance \ Covered}{Total \ time \ taken}$

= $\bf\dfrac{2x}{9x/200}$

= $\bf\dfrac{2x × 200}{9x}$

= $\bf\dfrac{400}{9}$

= 44.44 km/h

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Therefore, The average speed of whole journey is 44.44 km/h.