Physics, asked by hello9142, 7 months ago

1. An athlete completes one round of a circular track of diameter
Exercises
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 hl. On his return trip along the same
route, there is less traffic and the average speed is
30 km h-! 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 s-2 for 8.0 s. How far does the
boat travel during this time?
A driver for travellint 591​

Answers

Answered by vinithavt366
2

1.700 I think but I'm not sure about the answer

Answered by 10195
1

Answer:

1.  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.

2. Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms-1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms-1= 1.9 m/s

Average velocity while traveling from A to B =300/150 ms-1= 2 m/s

Average velocity while traveling from A to C =200/210 ms-1= 0.95 m/s

3. Distance traveled to reach the school = distance traveled to reach home = d (say)

Time taken to reach school = t1

Time taken to reach home = t2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t2= 30 kmph

Therefore, t1 = d/20 and t2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t1+t2)kmph = (d+d)/(d/20+d/30)kmph

= 120/5 kmh-1 = 24 kmh-1

4. Therefore, Abduls average speed for the entire trip is 24 kilometers per hour.Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms-2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at2

Therefore, the total distance traveled by boat in 8 seconds = 0 + 1/2 (3)(8)2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

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