Answer:
(a) Total north component = (26.89 + 25) = 51.89 metres
(b) Total west component = (-34.83 + 52) = 17.17 metres.
Explanation:
Do northwest components first.
North component = (sin 33.9 x 44) =26.89 metres.
West component = (cos 33.9 x 44) = -34.83metres.
Total west component = (-34.83 + 52) = 17.17 metres.
Total north component = (26.89 + 25) = 51.89 metres
(a) ![a_t = 0.800 m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%200.800%20m%2Fs%5E2)
The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:
![a_t = \frac{\Delta v}{\Delta t}](https://tex.z-dn.net/?f=a_t%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D)
This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is
![a_t = 0.800 m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%200.800%20m%2Fs%5E2)
(b) ![a_c = 0.9 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%200.9%20m%2Fs%5E2)
The centripetal acceleration component is given by
![a_c = \frac{v^2}{r}](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
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):
![a_c = \frac{(3.00 m/s)^2}{10.0 m}=0.9 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%20%5Cfrac%7B%283.00%20m%2Fs%29%5E2%7D%7B10.0%20m%7D%3D0.9%20m%2Fs%5E2)
(c) ![1.2 m/s^2, 48.4^{\circ}](https://tex.z-dn.net/?f=1.2%20m%2Fs%5E2%2C%2048.4%5E%7B%5Ccirc%7D)
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:
![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](https://tex.z-dn.net/?f=a%3D%5Csqrt%7Ba_t%5E2%2Ba_c%5E2%7D%3D%5Csqrt%7B%280.8%20m%2Fs%5E2%29%5E2%2B%280.9%20m%2Fs%5E2%29%5E2%7D%3D1.2%20m%2Fs%5E2)
and the direction is given by:
![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}](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%5Cfrac%7Ba_c%7D%7Ba_t%7D%3D%5Cfrac%7B0.9%20m%2Fs%5E2%7D%7B0.8%20m%2Fs%5E2%7D%3D1.125%5C%5C%5Ctheta%3Dtan%5E%7B-1%7D%281.125%29%3D48.4%5E%7B%5Ccirc%7D)
where the angle is measured with respect to the direction of the tangential acceleration.