Question

" A car is travelling around a horizontal circular track with radius r = 270 m at a constant speed v = 17 m/s as shown. The angle θA = 23° above the x-axis, and the angle θB = 58° below the x-axis.

1. What is the x component of the car’s acceleration when it is at point A?

2.What is the y component of the car’s acceleration when it is at point A?

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

4. What is the y component of the car’s acceleration when it is at point B?"

Answer 1

Given:
d=2 * pi * r
θA = 23°
θB = 58°
v = 17 m/s
r = 270 m
Using the formula:
Velocity = distance / time
So,substituting the values:

V = ( 2 * pi * r ) / t = 20.583 m/s

x component of velocity = sine ( 32 ° ) x 20.583
= 10.91 m/s

Answer 2

x component of car's acceleration when it is at point A is 0.554205452368

y component of car's acceleration when it is at point A is−0.915723236755

x component of car's acceleration when it is at point B is 0.892927967357

y component of car's acceleration when it is at point B is 0.590230781033

Explanation:

Here, given radius=270 m

& velocity = 17 m/s

Given angle:

`\theta_(A)=23^(\circ)\\\theta_(B)=58^(\circ)`

Now acceleration:

`a=(v^(2))/(r)\\=((17)^(2) )/(270)\\=(289)/(270)`

Now first we find the x component of car's acceleration when it is at point A

x component at A is
`a\cos(90^(\circ)-\theta_(A))`

=
`(289)/(270)\cos(90^(\circ)-23^(\circ))`

= −(-0.554205452368)

= 0.554205452368

Similarly, y component at A is
`a\sin(90^(\circ)-\theta_(A))`

=
`(289)/(270)\sin(90^(\circ)-23^(\circ))`

= −0.915723236755

Now, x component of car acceleration when it is at point B

x component is
`a\cos(90^(\circ)-\theta_(A))`

=
`(289)/(270)\cos(90^(\circ)-58^(\circ))`

= 0.892927967357

y component is
`a\sin(90^(\circ)-\theta_(A))`

=
`(289)/(270)\sin(90^(\circ)-58^(\circ))`

= 0.590230781033