A car is driving around a circular track .The radius of the track is 700.0 meter

Question

A car is driving around a circular track .The radius of the track is 700.0 meters. The car starts from rest at the eastern-most part of the track and starts speeding up in the counter-clockwise direction. After 20.0 seconds, it is going 60.0 mph (Not sure if miles or meters) in the counter-clockwise direction.

a) What is the tangential component of the car's acceleration? (Hint: its magnitude is constant, and its direction is in the direction the car is going.)

b) What is the the radial component of the car's acceleration at t = 15.0 seconds?

c) What is the acceleration vector (magnitude and direction) at t = 15.0 seconds? (Hint: you need to know where on the track the car is located at this instant.)

Explanation / Answer

1MPH =0.44704 m/s

V=60*0.44704 =26.82 m/s

a)

a=V/t =26.82/20

a=1.34 m/s^2

b)

V=Vo+at =0+1.34*15

V=20.16 m/s

a=V^2/r =20.16^2 /700

a=0.58 m/s^2

c)

a=sqrt(1.34^2+0.58^2[

a=1.46 m/s

theta =tan^-1(0.58/1.34)=23.4 degrees

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