Answer:
a.
Radial acceleration = 1.997 m/s^2
Tangential acceleration = 1.505 m/s^2
b.
It will take 35.6 sec to make one full circle.
Explanation:
Given:
A car travels in a flat circle of radius R.
Velocity of the car, v = 24 m/s west.
Total acceleration, a = 2.5 m/s^2 NW
a.
We have to find the radial [tex](a_c)[/tex] and tangential [tex](a_t)[/tex] components of acceleration of the car at that moment.
From the figure shown we can say that:
⇒ [tex]a_c=asin(\theta)[/tex] ⇒ [tex]a_t=acos(\theta)[/tex]
⇒ [tex]a_c=2.5\times sin(53)[/tex] ⇒ [tex]a_t=2.5\times cos(53)[/tex]
⇒ [tex]a_c=1.997\ m.s^-^2[/tex] ⇒ [tex]a_t=1.505\ m.s^-^2[/tex]
b.
Now we have to find the time taken for the angular displacement .
We know that:
⇒ [tex]a_c=\frac{v^2}{R}[/tex] and [tex]R=\frac{v^2}{a_c}[/tex] so,[tex]R=\frac{24^2}{1.997} = 288.433[/tex]
Let t be the time taken for the angular displacement θ to be 2 π .
2 π as it has taken full circle.
So,
⇒ [tex]\theta = \omega_i(t) + \frac{\alpha(t)^2}{2}[/tex] ...equation (i)
⇒ [tex]\theta=2\pi[/tex] , [tex]\omega_i =\frac{v}{R}[/tex] , [tex]\alpha=\frac{a_t}{R}[/tex] ...
⇒ [tex]\omega_i =\frac{v}{R}=\frac{24}{288.433} = 0.0832\ rad.s^-^1[/tex]
⇒ [tex]\alpha=\frac{a_t}{R}=\frac{1.505}{288.433} =0.00522\ rad.s^-^2[/tex]
⇒ Plugging all in equation (i)
⇒ [tex]\theta = \omega_i(t) + \frac{\alpha(t)^2}{2}[/tex]
⇒ [tex]2\pi = 0.08329(t) + \frac{0.00522(t)^2}{2}[/tex]
⇒ [tex]6.28 = 0.08329(t) + 0.00261(t)^2[/tex]
⇒ [tex]0.00261(t)^2+0.08329(t) -6.28=0[/tex]
⇒ Solving the above quadratic.
⇒ [tex]t=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex] where [tex]a=0.00261[/tex] , [tex]b=0.08329[/tex] and [tex]c=6.28[/tex]
⇒ [tex]t=\frac{0.08329 \pm \sqrt{(0.08329)^2-4\times 0.00261\times 6.28} }{2\times 0.00261}[/tex]
⇒ [tex]t=-67.5\ sec[/tex] and [tex]t=35.6\ sec[/tex]
⇒ Discarding the negative values.
⇒ time taken = 35.6 sec
Time taken by the car to make one full circle is 35.6 seconds.
