Answer

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**Hint:**In order to find the maximum velocity we will first find the time at which acceleration is zero and since acceleration is a derivative of velocity of a body with respect to time, using this concept we will solve for velocity at particular time. We will use the relation between velocity and acceleration which is $a = \dfrac{{dv}}{{dt}}$.

**Complete step by step answer:**

First, it’s given us that at $t = 0$ motorcycle was at rest at origin which means the value of $t = 0$ since, body is at origin. For the motorcycle to have a maximum velocity its acceleration must be zero and we will put acceleration magnitude zero in the given relation ${a_x}(t) = At - B{t^2}$ and we will get,

$At - B = 0$

$\Rightarrow t = \dfrac{A}{B}$

Putting the values $A = 1.50\,m\,{\sec ^{ - 2}}$ and $B = 0.120\,m\,{\sec ^{ - 4}}$ we get,

$t = 12.5\sec $

Now, acceleration can be written as:

$\dfrac{{dv}}{{dt}} = At - B{t^2}$

Or integrating both sides,

$\int {dv} = \int {(At - B{t^2})dt} $

$\Rightarrow v = A\dfrac{{{t^2}}}{2} + B\dfrac{{{t^3}}}{3}$

On putting the values of

$t = 12.5\sec $

$\Rightarrow A = 1.50\,m\,{\sec ^{ - 2}}$

$\Rightarrow B = 0.120\,m\,{\sec ^{ - 4}}$

In the equation $v = A\dfrac{{{t^2}}}{2} - B\dfrac{{{t^3}}}{3}$

We get,

$v = (0.75){(12.5)^2} - (0.04){(12.5)^3}$

$\therefore v = 39.1\,m\,{\sec ^{ - 1}}$

So when acceleration is zero the magnitude of the velocity of the motorcycle is maximum.

**Hence, the value of maximum velocity of the motorcycle is $v = 39.1\,m{\sec ^{ - 1}}$.**

**Note:**It should be remembered that, the basic integration formula of functions like $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $ and since motorcycle was at origin hence, the definite integral can be ignored since it was then when $t = 0$ and just at this time motorcycle attains its maximum velocity and also remember the relation between velocity and acceleration which is $a = \dfrac{{dv}}{{dt}}$.

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