Question

An automobile moves on a circular track of radius 1.04 km. It starts from rest from the point (x, y) = (1.04 km, 0 km) and moves counterclockwise with a steady tangential acceleration such that it returns to the starting point with a speed of 29.2 m/s after one lap. (The origin of the Cartesian coordinate system is at the center of the circular track.)

What are the car's position and velocity vectors when it is one-sixth of the way around the track? Express the vectors in terms of the unit vectors along the x- and y-axes.

r = ( _____)i + (_________)j km

v = (_______) i + (________)j m/s

What is the tangential acceleration at this point?

What is the centripetal acceleration at this point?

What is the least coefficient of static friction which will prevent the car from skidding in completing the first lap? (The answer needs to have at least 6 significant figures.)

Answer #1

d = 2 pi r = 2 x pi x 1040 m

v0 = 0

vf = 29.2 m/s

vf^2 - vi^2 = 2 a d

29.2^2 - 0^2 = 2(2 x pi x 1040)(a)

a_t = 0.0652 m/s^2 .....Ans(tangential acceleration)

at d = 2 x pi x 1040 / 6

v^2 - 0^2 = 2(2 x pi x 1040 / 6)(0.0652)

v = 11.92 m/s

at this time theta = 360/6 = 60 deg

r = 1.04 ( cos60 i + sin60 j)

r = 0.52i + 0.901 j km ....Ans

v = 11.92(cos(90+60)i + sin(60+90) j)

v = (-10.3 ) i + ( 5.96) j m/s ......Ans

centripetal acc. a = v^2 / r = 0.1366 m/s^2 .....Ans

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