ack of diametere covered and theExercises1. An athlete completes one r

1.	ack of diametere covered and theExercises1. An athlete completes one round of a circular track of diam200 m in 40 s. What will be the distance covered andisplacement at the end of 2 minutes 20 s?2. Joseph jogs from one end A to the other end B of a stra300 m road in 2 minutes 30 seconds and then turns aroand jogs 100 m back to point C in another 1 minute. What areJoseph's average speeds and velocities in jogging ta) from A toB and (b) from A to C?3. Abdul, while driving to school, computes the average speed forhis trip to be 20 km hl. On his return trip along the sameroute, there is less traffic and the average speed is30 km hl. What is the average speed for Abdul's trip?4. A motorboat starting from rest on a lake accelerates in a straightline at a constant rate of 3.0 ms for 8.0 s. How far does theboat travel during this time?5. A driver of a car travelling at 52 km h'' applies the brakes andaccelerates uniformly in the opposite direction. The car stopsin 5 s. Another driver going at 3 km h' in another car applieshis brakes slowly and stops in 10 s. On the same graph paper.plot the speed versus time graphs for the two cars. Which ofthe two cars travelled farther after the brakes were applied?6. Fig 8.11 shows the distance-time graph of three objects ABand C. Study the graph and answer the following questions:distance (km)0.4 0.8 1.2time (hour)1.6
Answer» 1. The diameter of a circular track is200 m200 mand one round is completed in40 s40 s. Circumference of a circular track is given as, C=2πrC=2πr. Where, the radius of circular track isrrand the circumference isCC. By substituting the given value in the above equation, we get C=2π×100= 200π mC=2π×100= 200π m So, the athlete covers a distance of200π m200π min40 s40 s. Total distance covered in2 min 20s2 min 20sor140 s140 sis, s=200π40×140≈2200 ms=200π40×140≈2200 m Since, every40 s40 sathlete covers one circular track therefore, athlete will take 3 round in120 s120 s. In the next20 s20 ssecond, he will cover half of the circumference. So, his displacement will be the diameter of the circular track that is200 m200 m. Thus, the distance covered in the given time is2200 m2200 mand the displacement is200 m200 m.

ack of diametere covered and theExercises1. An athlete completes one round of a circular track of diam200 m in 40 s. What will be the distance covered andisplacement at the end of 2 minutes 20 s?2. Joseph jogs from one end A to the other end B of a stra300 m road in 2 minutes 30 seconds and then turns aroand jogs 100 m back to point C in another 1 minute. What areJoseph's average speeds and velocities in jogging ta) from A toB and (b) from A to C?3. Abdul, while driving to school, computes the average speed forhis trip to be 20 km hl. On his return trip along the sameroute, there is less traffic and the average speed is30 km hl. What is the average speed for Abdul's trip?4. A motorboat starting from rest on a lake accelerates in a straightline at a constant rate of 3.0 ms for 8.0 s. How far does theboat travel during this time?5. A driver of a car travelling at 52 km h'' applies the brakes andaccelerates uniformly in the opposite direction. The car stopsin 5 s. Another driver going at 3 km h' in another car applieshis brakes slowly and stops in 10 s. On the same graph paper.plot the speed versus time graphs for the two cars. Which ofthe two cars travelled farther after the brakes were applied?6. Fig 8.11 shows the distance-time graph of three objects ABand C. Study the graph and answer the following questions:distance (km)0.4 0.8 1.2time (hour)1.6

Answer»

1. The diameter of a circular track is200 m200 mand one round is completed in40 s40 s.

Circumference of a circular track is given as,

C=2πrC=2πr.

Where, the radius of circular track isrrand the circumference isCC.

By substituting the given value in the above equation, we get

C=2π×100= 200π mC=2π×100= 200π m

So, the athlete covers a distance of200π m200π min40 s40 s.

Total distance covered in2 min 20s2 min 20sor140 s140 sis,

s=200π40×140≈2200 ms=200π40×140≈2200 m

Since, every40 s40 sathlete covers one circular track therefore, athlete will take 3 round in120 s120 s. In the next20 s20 ssecond, he will cover half of the circumference. So, his displacement will be the diameter of the circular track that is200 m200 m.

Thus, the distance covered in the given time is2200 m2200 mand the displacement is200 m200 m.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment