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:
\(\theta_{A}=23^{\circ}\\\theta_{B}=58^{\circ}\)
Now acceleration:
\(a=\frac{v^{2}}{r}\\=\frac{(17)^{2} }{270}\\=\frac{289}{270}\)
Now first we find the x component of car's acceleration when it is at point A
x component at A is \(a\cos(90^{\circ}-\theta_{A})\)
= \(\frac{289}{270}\cos(90^{\circ}-23^{\circ})\)
= −(-0.554205452368)
= 0.554205452368
Similarly, y component at A is \(a\sin(90^{\circ}-\theta_{A})\)
= \(\frac{289}{270}\sin(90^{\circ}-23^{\circ})\)
= −0.915723236755
Now, x component of car acceleration when it is at point B
x component is \(a\cos(90^{\circ}-\theta_{A})\)
= \(\frac{289}{270}\cos(90^{\circ}-58^{\circ})\)
= 0.892927967357
y component is \(a\sin(90^{\circ}-\theta_{A})\)
= \(\frac{289}{270}\sin(90^{\circ}-58^{\circ})\)
= 0.590230781033
