The car travels around the circular track having a radius of r = 200 m such that

Question

The car travels around the circular track having a radius of r = 200 m such that when it is at point A it has a velocity of 6 m/s, which is increasing at the rate of v = (0.06t) m/s^2, where t is in seconds. (Figure 1) Determine the magnitude of the velocity when it has traveled one-third the way around the track. Express your answer to three significant figures and include the appropriate units. Determine the magnitude of the acceleration when it has traveled one-third the way around the track. Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

distance travelled s = (1/3)*2*pi*r = (1/3)*2*pi*200 = 419 m

dv = integral a*dt

v - 6 = 0.06*t^2/2 = 0.03*t^2

v = 6 + 0.03*t^2

distance s = integral v*dt

s = 6t + 0.03t^3/3

s = 6t + 0.01*t^3

distance travelled s = (1/3)*2*pi*r = (1/3)*2*pi*200 = 419 m

(1/3)*2*pi*200 = 6t + 0.01t^3

t = 29.03 s

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part(A)

magnitude of velocity

v = 6 + (0.03*29^2)

v = 31.2 m/s <<<=========answer

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part B

radial accelration ar = v^2/r = 31.2^2/200 = 4.87 m/s^2

tangentail acceleration at = 0.06*29 = 1.74 m/s^2

total accelration a = sqrt(4.87^2+1.74^2) = 5.1 m/s^2     <<<<++++++++++ANSWER

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