Question

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 the point C in another 1 minute. What are Joseph's average speeds and velocities in jogging :
(a.) from A to B?
(b.) from A to C?

Answer 1

Answer:

Explanation

From point A to B

Distance covered=300m

Displacement=300m

Time taken=2 min 30 sec=(2×60)+30=150 sec

Average speed=  

Time-taken

Distance covered  =  150

300 =2 m/sec

Average velocity=  

Time-taken

Displacement  =  150

300  =2 m/sec

From point A to C

Distance covered=300+100=400m

Displacement=300−100=200m

Time taken=3 min 30 sec=(3×60)+30=210 sec

Average speed=  Time-taken

Distance covered  =  210

400 =1.90 m/sec

Average velocity=Time-taken

Displacement  =  210

200 =0.95 m/sec

Answer 2

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Given, diameter of the track (d) = 200m

Therefore, circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of laps completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

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