Answer to Question #108999 in Physics for Tumi

Question

Determine the velocity of a point, it’s tangential and normal accelerations in 10 seconds after motion begins if at the moment 3 seconds it’s normal acceleration was 0.9m/s^2. The equation of point motion is S=120+7t+2t^2m

Answer 1

$$v(t)=\frac{d}{dt}s=7+2(2t)=7+4t$$

Thus, the velocity of a point in 10 seconds after motion begins

$$v(10)=7+4(10)=47\frac{m}{s}$$

$$a_t(t)=\frac{d}{dt}v=4$$

Thus, it’s tangential acceleration in 10 seconds after motion begins

$$a_t(10)=4\frac{m}{s^2}$$

$$a_n(3)=\frac{v(3)^2}{r}=\frac{(7+4(3))^2}{r}=0.9\frac{m}{s^2}$$

$$r=401\ m$$

Thus, it’s normal acceleration in 10 seconds after motion begins

$$a_n(10)=\frac{(7+4(10))^2}{401}=5.5\frac{m}{s^2}$$