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?

1 Answer
Jul 15, 2018

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]}