(a) [tex]a_t = 0.800 m/s^2[/tex]
The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:
[tex]a_t = \frac{\Delta v}{\Delta t}[/tex]
This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is
[tex]a_t = 0.800 m/s^2[/tex]
(b) [tex]a_c = 0.9 m/s^2[/tex]
The centripetal acceleration component is given by
[tex]a_c = \frac{v^2}{r}[/tex]
where
v is the tangential speed
r is the radius of the trajectory
When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):
[tex]a_c = \frac{(3.00 m/s)^2}{10.0 m}=0.9 m/s^2[/tex]
(c) [tex]1.2 m/s^2, 48.4^{\circ}[/tex]
The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:
[tex]a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.8 m/s^2)^2+(0.9 m/s^2)^2}=1.2 m/s^2[/tex]
and the direction is given by:
[tex]tan \theta =\frac{a_c}{a_t}=\frac{0.9 m/s^2}{0.8 m/s^2}=1.125\\\theta=tan^{-1}(1.125)=48.4^{\circ}[/tex]
where the angle is measured with respect to the direction of the tangential acceleration.
