The acceleration of a motorcycle is given by ax(t)=at−bt2, where a=1.50m/s3 and b=0.120m/s4. the motorcycle is at rest at the origin at time t=0. calculate the maximum velocity it attains.

From given information, the acceleration is
a(t) = 1.5t - 0.12t²  m/s²

Integrate to obtain the velocity.
v(t) = (1/2)*1.5t² - (1/3)*0.12t³ + c₁   
      = 0.75t² - 0.04t³ + c₁  m/s

Because v(0) = 0 (given), therefore c₁ = 0
The velocity is
v(t) = 0.75t² - 0.04t³  m/

The velocity is maximum when the acceleration is zero. That is,
t(1.5 - 0.12t) = 0
t = 0 or t = 1.5/.12 = 12.5 s
Reject t = 0 because it yields zero value.

The maximum velocity is
v(12.5) = 0.75*(12.5²) - 0.04*(12.5³) = 39.0625 m/s

The maximum velocity is 39.06 m/s (nearest hundredth)

The graph shown below displays the velocity.