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
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