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

Question

6. 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 in the counter-clockwise direction. Draw a picture, it will help. 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 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

Radius of the track: R = 700 m, initial speed: u = 0

After t = 20 s, v = 60 mph = 26.82 m/s

(a) Suppose the tangetial acceleration = a(t)

Using linear motion equation,

     v = u + a*1

=> v = u + a(t)*t

=> 26.82 = 0 + a(t)*20 =>   a(t) = 26.82/20 = 1.34 m/s^2

(b) For t = 15 s,

v = u + a(y)*t = 0 + 1.34*15 = 20.12 m/s

The radial acceleration at this time, a(r) = v^2 / R = 20.12^2 / 700 = 0.58 m/s^2

(c) At t = 15 s,

a(t) = 1.34 m/s^2 and a(r) = 0.58 m/s^2

Magnitude of acceleration vector  = sqr{ a(t)^2 + a(r) } =   sqr{ 1.34^2 + 0.58 } = 1.46 m/s^2

For the direcion, = tan^-1{ a(t) / a(r) }

   = tan^-1(1.34 / 0.58)

= tan^-1(2.31) = 67° ......This is the angle from the line joining position and center.

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