91Ó°ÊÓ

The given data can be listed below as,

• The value of A is $1.50 \mathrm{m}/{\mathrm{s}}^3$.
• The value of B is .$0.120 \mathrm{m}/{\mathrm{s}}^4$
• The motorcycle rests at the origin at the time t=0.

This law states that a body will continue to be in rest or in motion unless it is acted by an external force.

Differentiating the equation of motion along a straight line gives the velocity and integrating the velocity gives the position as the function of time. Moreover, with the help of the gathered velocity, the maximum velocity can be attained.

a) The given equation of the acceleration of a motorcycle can be expressed as:

$a_x\left(t\right)=At-B{t}^2$… i)

Here,$a_x\left(t\right)$ is the acceleration of the motorcycle with time, A and B are the constants and t are the time taken.

We know that,$a=\frac{dv}{dt}$, hence, from equation i), we get-

$$\begin{gathered} a=\frac{dv}{dt} \\ ∫dv=a∫dt \\ v=∫\left(At−B{t}^2\right)dt \\ v=\left(\frac{A{t}^2}{2}−\frac{B{t}^3}{3}\right) \end{gathered}$$… ii)

We also know that, $v=\frac{dx}{dt}$, hence, from equation ii), we get-

$$\begin{gathered} v=\frac{dx}{dt} \\ ∫dx=∫vdt \\ x=∫\left(\frac{A{t}^2}{2}−\frac{B{t}^3}{3}\right)dt \\ x=\frac{A{t}^3}{6}−\frac{B{t}^4}{12} \end{gathered}$$

Thus, the position and the velocity as a function of time are$\frac{A{t}^3}{6}-\frac{B{t}^4}{12}$and$\left(\frac{A{t}^2}{2}-\frac{B{t}^3}{3}\right)$respectively.

b) As we know that,$a=\frac{dv}{dt}$which is 0, hence, we get-

$$\begin{gathered} \frac{2At}{2}−\frac{3B{t}^2}{3}=0 \\ At−B{t}^2=0 \\ t=0\text{and} \\ t=\frac{A}{B} \end{gathered}$$

As t=0 is negligible, the using the second value of t, we get-

$$\begin{gathered} V_{a_{\mathrm{t}}\mathrm{at}=\frac{A}{B}}=Aâ‹\dots {\left(\frac{A}{B}\right)}^2â‹\dots \frac{1}{2}−\frac{B}{3}â‹\dots {\left(\frac{A}{B}\right)}^3 \\ =\frac{A^3}{2B^2}−\frac{A^3}{3B^2} \end{gathered}$$… iii)

Hence, we can get

Substituting the value of A and B is the equation iii), we get-

$$\begin{gathered} V_{max}=\frac{{\left(1.50\mathrm{m}/{\mathrm{s}}^3\right)}^3}{2{\left(0.12\mathrm{m}/{\mathrm{s}}^4\right)}^2}−\frac{{\left(1.50\mathrm{m}/{\mathrm{s}}^3\right)}^3}{3{\left(0.12\mathrm{m}/{\mathrm{s}}^4\right)}^2} \\ {V}_{max}=39.06\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the maximum velocity attained by the motorcycle is 39.06 m/s.