A particle moves in a circular path at constant speed. The initial location of t

Question

A particle moves in a circular path at constant speed. The initial location of the particle is (4.07m,5.91m) at t=3.72 s. The velocity of the particle at this time is 2.81m/sunit vector j.png(i.e., directed along the positive y-axis). Acceleration at this time is along the +x axis. At t=10.47 s, v is -2.81m/sunit vector i.pngand the acceleration is directed along the +y axis. Find the radius of the circle (in meters). Enter only the numerical part in the answer box. For this problem assume the time interval is less than one full period.

Explanation / Answer

Given,

v1 = 2.81 m/s j ; v2 = -2.81 m/s i

t1 = 3.72 ; t2 = 10.47 s

a = 2.81 - (-2.81)/(10.47 - 3.72) = 0.83 m/s^2

we know that,

a = v^2/R => R = v^2/a

R = 2.81^2/0.83 = 9.51 m

Hence, R = 9.51 m

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