Question

A motorcycle drive from a to b with a uniform speed of 40 km h and returned back with a speed of 50 km h. Find average speed and average velocity

Answer 1

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✦ Required Answer:

♦️ GiveN:

• Motorcycle drives a particular distance and returns back to the same point.
• While going, it's speed was 40 km/hr and while returning, it's speed was 50 km/hr.

♦️ To FinD:

• Average speed....?
• Average velocity .....?

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✦ Explanation of Concept:

The above question has provided us the distance as well as displacement. Also, the speed and velocity for particular distance is also given. From here, we can find average speed and average velocity by using formulae,

$\large{\underline{\underline{\rm{\red{Average\: speed}}}}}$

$\large{ \boxed{ \rm{avg. \: speed = \frac{Total \: distance}{Total \: time}}}}$

$\large{\underline{\underline{\rm{\red{Average\: velocity}}}}}$

$\large{ \boxed{ \rm{Avg. \: velocity = \frac{Total \: distplacement}{Total \: time}}}}$

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✏By using these formulae, we can solve this question.

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✦ Solution:

Let the distance from A to B be s, while returning it again covers the same distance s

Then, total distance = 2s

Now, Speed from A to B = 40 km/hr

$\large{ \rm{then \: Time = \frac{Distance}{Speed}}} \\ \large{ \rm{Time = \frac{s}{40} \: hr }}$

Also, given speed from B to A = 50 km/hr

$\large{ \rm{Time = \frac{s}{50} \: hr}} \\ ..........................$

By using formula,

$\large{ \rm{ \dashrightarrow \: Avg. \: speed = \frac{s + s}{ \frac{s}{40} + \frac{s}{50} } \: km {h}^{ - 1} }} \\ \\ \large{ \rm{ \dashrightarrow \: Avg. \: speed = \frac{2s}{ \frac{5s + 4s}{200} } \: km {h}^{ - 1} }} \\ \\ \large{ \rm{ \dashrightarrow \: Avg. \: speed = \cancel{ \frac{2s \times 200}{9s} km {h}^{ - 1} }}} \\ \\ \large{ \rm{ \dashrightarrow \: Avg. \: speed \approx \: \boxed{ \rm{ \red{44.45 \: km {h}^{ - 1} }}}}}$

Hence, Average speed = 44.45 km/hr

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Now, as we can see that the object started from A and ends his journey in A itself. This means that the initial and final point are same. Thus, displacement is o because displacement is the shortest path between initial and final point. Hence, Total Displacement = 0

$\large{ \rm{ \dashrightarrow \: Avg. \: velocity \: = \boxed{ \rm{ \red {0}}}}}$

Hence, Average velocity = 0 km/hr.

$\large{ \therefore{ \underline{ \underline{ \rm{ \purple{Hence,\: solved \: \dag}}}}}}$

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Answer 2

★ DIAGRAM:

A B

.________bike_____________.

Note: The bolded text is the motorcycle

★ GIVEN:

Motorcycle travels from A to B with a speed of 40km/h

And returning with a speed of 50km/h

★ TO FIND:

Average speed of motor cycle

★ SOLUTION:

Let , distance travelled from A - B be x

Now, total distance = A to B + B to A = x + x = 2x

Finding total time

Using

Time ( t₁ ) = Distance/Speed

Time ( t₁ ) = x/40

Time ( t₂ ) = x/50

Total time = t₁ + t₂

Total time = x/40 + x/50

Total time = 9x/20

________________________

Avg speed = Total distance/total time

Avg speed = 2x/9x/20

Avg speed = 2 × 20

Avg speed = 40km/h .…… ( Answer )