Physics, asked by summu26, 8 months ago

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​

Answers

Answered by miraculousladybug18
1

2 by one fourth

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Answered by yagnasrinadupuru
3

ANSWER

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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