Question

A car runs at a constant speed on a circular track of radius 150m taking 62.8s on each leap. The average speed and average velocity on each complete lap is? (π=3.14)​

Answer 1

Answer:

$\underline{\purple{\ddot{\Mathsdude}}}$

Given:-

• A car runs at a constant speed on a circular track of radius 150m taking 62.8sec

To prove:-

• average speed
• average velocity

Proof:-

• At first we should need to find circumference of the circle

Circumference of the circle = 2 pi r

= 2×22/7×150

=942m.

Now ,

Average speed =total distance/ time

= 942/62.8

= 15m/s

Therefore , starting point and ending point is same.

• From this we can know that average velocity = 0m/s

Hopeit helps u mate .

Thank you .

Answer 2

Answer:

$\pink{\sf\: \odot \sf \: \: average \: speed \: \: \bar{v} = 15 \: m{s}^{ - 1} } \\ \orange {\sf \: \odot \sf \: \: average \: velocity \: \: v = 0 \: m {s}^{ - 1} }$

Explanation:

Given,

$\odot \sf \: radius \: of \: a \: circular \: path \: \: r = 150 \: m \\ \odot \sf \: Circumference \: \: d = 2\pi \: r \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \times 3.14 \times 150 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 942 \: m \\ \\ \pink{ \therefore \sf\: distance \: d \: = 942 \: m} \\ \\ \odot \sf \: time \: \: t = 62.8 \: d$

To find :

1) average speed = $\bar{v}$

2) average velocity

Solution:

1)

$\bf \:w e \: know \: that \: \\ \sf \: average \: speed \: \: \bar{v} = \frac{total \: distance(d)}{total \: time(t)} \\ \therefore \: \sf \bar{v} = \frac{d}{t} \\ \: \: \: \: \sf \: \: \: \: = \frac{942}{62.8} \\ \sf \: \: \: \: \: \: \: \: = 15 \: m {s}^{ - 1} \\ \\ \pink{\sf\therefore \: average \: speed \: \: \bar{v} = 15 \: m{s}^{ - 1} }$

2)

Here, initial point and final point are same.

So,

Displacement s = 0 m

cause,

displacement is a difference between final and initial point.

Now,

$\sf \: average \: velocity \: \: v = \frac{total \: displacement(s)}{total \: time(t)} \\ \therefore \sf \: \: v = \frac{s}{t} \\ \: \: \: \: \: \: \: \: \: = \frac{0 }{62.8} \\ \: \: \: \: \: \: \: \: \: = 0 \: m {s}^{ - 1} \\ \\ \orange {\sf\therefore \: average \: velocity \: \: v = 0 \: m {s}^{ - 1} }$