Quest Academy

Here we have, Diameter = 200 m, therefore, radius = 200m/2 = 100 m Time of one rotation = 40s Time after 2m20s = 2 x 60s + 20s = 140s Distance after 140 s = ? Displacement after 140s =? Motion Ex1 Motion Ex2 (a) Distance after 140s We know that,distance=velocity ×time ?distance=15.7m/s ×140 s = 2198 m (b) Displacement after 2 m 20 s i.e. in 140 s Since,rotatin in 40 s=1 Motion Ex3 Therefore, in 3.5 rotations athlete will be just at the opposite side of the circular track, i.e. at a distance equal to the diameter of the circular track which is equal to 200 m. Therefore, Distance covered in 2 m 20 s = 2198 m And, displacement after 2 m 20 s = 200m 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? Answer:- Motion Ex4 Here we have, Distance from point A to B = 300 m Time taken = 2 minute 30 second = 2 x 60 + 30 s = 150 s Distance from point B to C = 100 m Time taken = 1 minute = 60 s (a) Average speed and velocity from point A to B Motion Ex5 (b) Average speed and velocity from B to C Motion Ex6 Therefore,average velocity=1.66 m/s west