Centripetal acceleration of a cyclist complet | vedclass

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Question

(A) The cyclist completes $7$ rounds in $1$ minute $(60 \, s)$.The circumference of the circular track is $C = 2 \pi r = 2 	imes 3.14159 	imes 5 \,

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(A) The cyclist completes $7$ rounds in $1$ minute $(60 \, s)$.The circumference of the circular track is $C = 2 \pi r = 2 	imes 3.14159 	imes 5 \, m \approx 31.416 \, m$.The total distance covered in $7$ rounds is $D = 7 	imes 31.416 \, m = 219.912 \, m$.The constant speed $v$ is given by $v = 	ext{Distance} / 	ext{Time} = 219.912 \, m / 60 \, s \approx 3.665 \, m/s$.The centripetal acceleration $a_c$ is given by $a_c = v^2 / r$.Substituting the values: $a_c = (3.665)^2 / 5 \approx 13.432 / 5 \approx 2.686 \, m/s^2$.Rounding to one decimal place,we get $2.7 \, m/s^2$.Thus,Option $A$ is correct.

Centripetal acceleration of a cyclist completing $7$ rounds in a minute along a circular track of radius $5 \, m$ with a constant speed,is ......... $m/s^2$.

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