Question

The acceleration of a motorcycle is given by  a_x(t)=At-Bt², where A=1.50m/s³  and B=0.12m/s⁴ . The motorcycle is at rest at the origin at time t=0.

a. Find its position and velocity as a function of time.

b. Calculate the maximum Velocity

 

Answer 1

(a) We can find the velocity of the motorcycle by taking the integral of the acceleration over time:

$$v(t)=\int a(t)dt,$$ $$v(t)=\int (At-Bt^2)dt=\dfrac{At^2}{2}-\dfrac{Bt^3}{3}.$$

We can find the position of the motorcycle by taking the integral of the velocity over time:

$$s(t)=\int v(t)dt,$$ $$s(t)=\int (\dfrac{At^2}{2}-\dfrac{Bt^3}{3})dt=\dfrac{At^3}{6}-\dfrac{Bt^4}{12}.$$

(b) Let's first take the derivative from the velocity with respect to time:

$$\dfrac{d}{dt}(\dfrac{At^2}{2}-\dfrac{Bt^3}{3})=At-Bt^2.$$

Let's set the derivative equal to zero and find the time at which the velocity of the motorcycle is maximum:

$$At-Bt^2=0,$$ $$t(A-Bt)=0.$$

This equation has two roots: $t=0$ and $t=\dfrac{A}{B}.$ Since, at $t=0$ the velocity of the motorcycle is zero, its velocity is maximum at $t=\dfrac{A}{B}$.

Substituting $t$ into the equation for $v(t)$ we get:

$$v_{max}=\dfrac{A(\dfrac{A}{B})^2}{2}-\dfrac{B(\dfrac{A}{B})^3}{3}=\dfrac{A^3}{6B^2}.$$

Let's substitute the numbers:

$$v_{max}=\dfrac{(1.50\ \dfrac{m}{s^3})^3}{6\cdot(0.12\ \dfrac{m}{s^4})^2}=39.1\ \dfrac{m}{s}.$$