A car is traveling around a horizontal circular track with radius r = 210 m as s

Question

A car is traveling around a horizontal circular track with radius r = 210 m as shown. It takes the car t = 51 s to go around the track once. The angle A = 31° above the x axis, and the angle B = 55° below the x axis.

1)

What is the magnitude of the car’s acceleration?

m/s2

2)

What is the x component of the car’s velocity when it is at point A

m/s

3)

What is the y component of the car’s velocity when it is at point A

m/s

4)

What is the x component of the car’s acceleration when it is at point B

m/s2

5)

What is the y component of the car’s acceleration when it is at point B

m/s2

6)

As the car passes point B, the y component of its velocity is

increasing

constant

decreasing

Explanation / Answer

radius r = 210 m

time, t = 51 s

angle produced at A, A = 31°

angle produced at B, B = 55°

velocity=distance/time

We know that the distance is the cirumference of the circle, which is nothing but 2 * pi* r

So,velocity=25.86 m/s

x component of velocity at A is the product of velocity and sin of the angle produced along A,

So, vx=25.86*sin 31=13.32 m/s

x component of velocity at B is the product of velocity and sin of the angle produced along B,

So, vb=25.86*sin 55=21.18 m/s

y component of velocity at A is the product of velocity and cos of the angle produced along A,

So, vy=25.86*cos 31=22.16 m/s

acceleration=v2/r=(25.86)2/210=3.18 m/s2

x component of acceleration at A is the product of accelaration and sin of the angle produced along A,

So, ax=3.18*sin 31=1.628 m/s2

x component of velocity at B is the product of accelaration and sin of the angle produced along B,

So, ax=3.18*sin 55=2.604 m/s2

y component of accelaration is,

ay=3.18*cos 55=1.823 m/s2

It decreases

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