Question

25. An athlete completes one round of a circular track of radius 20m R in 40 sec. Find the magnitude of
displacement (in m) of the athlete at the end of 2min 20 sec.​

Answer 1

Explanation:

HEY MATE ..........

$the \: radius \: of \: circular \: track \: is \: \\ \\ given \: as \: 20 |m| \\ \\ in \: 40 |sec| the \: athlete \: complete \: one \: round \: \\ \\ so \: in \: 2min \: and \: 20sec \: that is \: 140sec \: \\ \\ the \: athlete \: will \: complete \: \\ \\ = \frac{140}{40} = 3.5(three \: and \: a \: half \: rounds) \\ \\ one \: round \: is \: considered \: \\ \\ as \: the \: circumference \: of \: the \: circular \: track \: \\ \\ the \: distance \: covered \: in \: 140sec \: \\ \\ = 2\pi \times 3.5 = 2 \times 3.14 \times 100 \times 3.5 = 2200 |m| \\ \\ for \: each \: complete \: round \: \\ \\ the \: displacement \: is \: zero \: \\ \\ therefore \: \\ for \: 3 \: complete \: rounds \: \: \\ \\ the \: displacement \: will \: be \: zero \: \\ \\ at \: the \: end \: of \: his \: motion \: \\ \\ the \: athlete \: will \: be \: in \: the \: \\ \\ diametrically \: opposite \: position \: \\ \\ that \: is \: displacement \: = radius = 20 |m| \\ \\ hence \: the \: distance \: covered \: is2200 |m | and \: the \: displacement \: is \: 20 |m|$

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