Question

. A motorcyclist drives from A to B with a uniform speed of 40 km/h and returns back with a speed of 50 km/h. Find his average speed and average velocity

Answer 1

Let us assume that the motorcyclist covered the 'd' distance.

A motorcyclist drives from A to B with a uniform speed of 40 km/h.

Time = Distance/Speed

t1 = d/40 hr

Motorcyclist returns back with a speed of 50 km/h.

t2 = d/50 hr

Total time taken by motorcyclist = t1 + t2

= d/40 + d/50

= (5d + 4d)/200

= 9d/200

Total distance covered is 2d km. Since, the motorcyclist starts from A to B and then returns back using the same pathway.

Average speed is defined as the ratio of total distance covered with respect to total time taken.

Average speed = 2d/(9d/200)

= (2d × 200)/9d

= 44.44

Therefore, the average speed of the motorcyclist is 44.44 km/hr.

On the other hand, average velocity is defined as the ratio of total displacement with respect to total time taken.

Since, the initial and final positions/pounts of the motorcyclist is same. So, the displacement is zero.

Average velocity = 0/(9d/200)

= 0

Therefore, the average velocity of the motorcyclist is 0 km/hr.

Answer 2

Answer:

avg.velocity=0

avg.speed=44.44km/hr

Explanation:

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$avg.velocity = \frac{final \: position - initial \: position}{final \: time - initial \: time}$

here,the cyclist drives to B but returns back to A itself. so,

$avg.velocity = \frac{a - a}{t2 - t1} \\ \: \: = 0$

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$avg.speed = \frac{2v1v2}{v1 + v2}$

This equation is only applicable if both distances are same, (here, bike travels from A to B and returns to A, so we can use this eqtn)

therefore,

$avg.speed = \frac{2 \times 40 \times 50}{40 + 50} \\ = \frac{400}{9} \\ = 44.4$

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hope it helps...!!!