Question

A motocycle starts from rest at t = 0.0 on a circular track

of400 meter radious. The tangential component of acceleration
isat=2+0.2t m/sec^2. At time t =10 sec determine
thedistance the motorcycle has moved along the track and the
magnitudeof its acceleration. (Knowing that a=at
+an).

Answer 1

`S_(10)=133.3m\\a_(10)=4.589m/s^(2)`

Step-by-step 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:

`V(t) = \int {a(t)} \, dt =  \int {2+0.2*t} \, dt = 2*t + 0.2*(t^(2))/(2)`

We will integrate once again to get distance:

`S(t) = \int {V(t)} \, dt = \int {2*t+0.2*(t^(2))/(2) } \, dt  = t^(2)+0.2*(t^(3))/(6)`

Now we just need t evaluate S(10s):

S(10) = 133.3m

For the acceleration, we know that:

`a = \sqrt{a_(t)^(2)+a_(N)^(2)}`

where
`a_(t)(10)=2+0.2*(10)=4m/s^(2)`

and
`a_(N)(10)=(V(10)^(2))/(R)=2.25m/s^(2)`

So, finally:

`a = \sqrt{a_(t)^(2)+a_(N)^(2)} = 4.589m/s^(2)`