Question

The acceleration of a motorcycle is given by...?

The acceleration of a motorcycle is given by a=At-Br^2 where A=1.20 m/s^3 and B=0.120 m/s^3. It is at rest at the origin at time t=0 s.

a.) What is it's position and velocity as functions of time?
b.) What's the maximum speed it attains?

Answer 1

Please see the explanation below

Explanation:

The acceleration is

`a(t)=At-Bt^2`

The velocity is the integral of the acceleration

`v(t)=inta(t)dt=int(At-Bt^2)dt`

`=At^2/2-Bt^3/3+C`

Plugging in the initial conditions

`v(0)=0`

Therefore,

`0=0-0+C`

`C=0`

So,

The velocity is

`v(t)=1.2/2t^2-0.12/3t^3=0.6t^2-0.04t^3`

The position is the integral of the velocity

`p(t)=intv(t)dt=int(0.6t^2-0.04t^3)dt`

`p(t)=0.6t^3/3-0.04t^4/4+C_1`

Plugging in the initial conditions

`p(0)=0-0+C_1=0`

Therefore,

The position is

`p(t)=0.2t^3-0.01t^4`

The maximum speed is when

`(dv)/(dt)=0`

`1.2t-0.12t^2=0`

`=>`, `t(1.2-0.12t)=0`

Therefore,

`{(t=0),(t=1.2/0.12=10s):}`

The maximum velocity is `v_(max)=60-40=20ms^-1`

See the graph of velocity v/s time

graph{0.6x^2-0.04x^3 [-4.93, 52.82, -2.02, 26.85]}