a cyclist travels from centre of a circular Park of radius 1 kilometre and reaches a point P after cycling along one fourth of the circumference along p q he returns to the centre of the park along Q if the total time taken is time minutes calculate the net displacement average velocity and average speed of the cyclist
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2 by one fourth
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ANSWERSince the initial position coincides with the final position. So, the net displacement of the cyclist = zero
ANSWERSince the initial position coincides with the final position. So, the net displacement of the cyclist = zeroAverage speed of the cyclist =
ANSWERSince the initial position coincides with the final position. So, the net displacement of the cyclist = zeroAverage speed of the cyclist = Total time taken/Total distance travelled
=
= 10
= 10OP+PQ+QO
= 10OP+PQ+QO
= 10OP+PQ+QO km min
= 10OP+PQ+QO km min −1
= 10OP+PQ+QO km min −1
= 10OP+PQ+QO km min −1 =
= 10OP+PQ+QO km min −1 = 20
= 10OP+PQ+QO km min −1 = 20π+4
= 10OP+PQ+QO km min −1 = 20π+4
= 10OP+PQ+QO km min −1 = 20π+4 ×60kmh
= 10OP+PQ+QO km min −1 = 20π+4 ×60kmh −1
= 10OP+PQ+QO km min −1 = 20π+4 ×60kmh −1 =21.4kmh
= 10OP+PQ+QO km min −1 = 20π+4 ×60kmh −1 =21.4kmh −1
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