Question

1. An athlete completes one round of a circular track of diameter
200 m in 40 s. What will be the distance covered and the
displacement at the end of 2 minutes 20 s?
2. Joseph jogs from one end A to the other end B of a straight
300 m road in 2 minutes 30 seconds and then turns around
and jogs 100 m back to point C in another 1 minute. What are
Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h-!. On his return trip along the same
route, there is less traffic and the average speed is 30 km h-1. What is the average speed for Abdul's trip?
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s2 for 8.0 s. How far does the
boat travel during this time?​

Answer 1

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Answer 2

Answer:

1) The diameter of the circular track is given as 200 m. That is, 2r=200 m.

From this radius is calculated as Radius, r = 100 m.

In 40 sec the athlete complete one round. So, in 2 mins and 20 secs, that is, 140 sec the athlete will complete = 140 / 40 = 3.5 (three and a half) rounds.

One round is considered as the circumference of the circular track.

The distance covered in 140 sec = 2πr×3.5=2×3.14×100×3.5=2200 m.

For each complete round the displacement is zero. Therefore for 3 complete rounds, the displacement will be zero.

At the end of his motion, the athlete will be in the diametrically opposite position. That is, displacement = diameter = 200 m.

Hence, the distance covered is 2200 m and the displacement is 200 m.

2)  The answer 2 is attached in the image.

3) Method I:

Let the distance traveled by Abdul from home to school = s km

time taken to reach the school=t1 sec

For Return journey , Abdul cover distance =s km

time=t2 sec

∴ Average speed for forward journey[home - school ] = Total distance/ Total time.

20 km/h =s/t1

∴t1=s/20 h ------eq(1)

Average speed for backward journey[school -home] = Total distance/ Total time.

30 km/h =s/t2

t2=s/30h -----eq(2)

Average distance for entire journey = Total distance/ Total time

                                                        =(s+s)/[s/20 +s/30]

                                                        =2s/s[1/20+1/30]

                                                         =2x20x30/50

                                                          =24 km/hr

Shortcut method :

If equal distance are covered with speeds v1 and v2 then average distance=2v1v2/v1+v2

                          = 2x20x30/50

                           =24km/hr

∴ The average speed for Abdul's trip is 24km/h

4) The answer is attached in the image.

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