Answer:
[tex]S_{10}=133.3m\\a_{10}=4.589m/s^{2}[/tex]
Explanation:
In order to know the value of the speed at any time t, we need to integrate the acceleration. Once we get the speed vs time, we need to integrate again to get the distance traveled by the motorcycle vs time. So let's start with the speed first:
[tex]V(t) = \int {a(t)} \, dt = \int {2+0.2*t} \, dt = 2*t + 0.2*\frac{t^{2}}{2}[/tex]
We will integrate once again to get distance:
[tex]S(t) = \int {V(t)} \, dt = \int {2*t+0.2*\frac{t^{2}}{2} } \, dt = t^{2}+0.2*\frac{t^{3}}{6}[/tex]
Now we just need t evaluate S(10s):
S(10) = 133.3m
For the acceleration, we know that:
[tex]a = \sqrt{a_{t}^{2}+a_{N}^{2}}[/tex]
where [tex]a_{t}(10)=2+0.2*(10)=4m/s^{2}[/tex]
and [tex]a_{N}(10)=\frac{V(10)^{2}}{R}=2.25m/s^{2}[/tex]
So, finally:
[tex]a = \sqrt{a_{t}^{2}+a_{N}^{2}} = 4.589m/s^{2}[/tex]
