" 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?"

A car is travelling around a horizontal circular track with radius r 270 m at a constant speed v 17 ms as shown The angle θA 23 above the xaxis and the angle θ class=

Respuesta :

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:

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


Step-by-step explanation:

Here, given radius=270 m

             & velocity = 17 m/s

Given angle:

[tex]\theta_{A}=23^{\circ}\\\theta_{B}=58^{\circ}[/tex]

Now acceleration:

[tex]a=\frac{v^{2}}{r}\\=\frac{(17)^{2} }{270}\\=\frac{289}{270}[/tex]

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

x component at A is [tex]a\cos(90^{\circ}-\theta_{A})[/tex]

                              = [tex]\frac{289}{270}\cos(90^{\circ}-23^{\circ})[/tex]

                              = −(-0.554205452368)

                              = 0.554205452368

Similarly, y component at A is [tex]a\sin(90^{\circ}-\theta_{A})[/tex]

                              = [tex]\frac{289}{270}\sin(90^{\circ}-23^{\circ})[/tex]

                              = −0.915723236755

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

x component is  [tex]a\cos(90^{\circ}-\theta_{A})[/tex]

                              = [tex]\frac{289}{270}\cos(90^{\circ}-58^{\circ})[/tex]

                              = 0.892927967357

y component is [tex]a\sin(90^{\circ}-\theta_{A})[/tex]

                              = [tex]\frac{289}{270}\sin(90^{\circ}-58^{\circ})[/tex]

                              = 0.590230781033