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]}