The acceleration of a bus is given by \(a_{x}(t)= \alpha t\), where \(\texttip{\alpha}{alpha}\) = 1.10\({\rm m/s^3}\) is a constant.
To find the displacement and velocity of the bus at a certain time, we need to integrate the given acceleration with respect to time.
The acceleration function is \(a_x(t) = \alpha t\), where \(\alpha\) is given as 1.10 m/s\(^3\).
Integrating the acceleration function will give us the velocity function.
\(v_x(t) = \int a_x(t) \, dt\)
Since \(a_x(t) = \alpha t\), we can rewrite the equation as:
\(v_x(t) = \int \alpha t \, dt\)
To integrate \(t\) with respect to \(dt\), we use the power rule of integration:
\(\int t \, dt = \frac{1}{2}t^2 + C\)
In this case, we don't have a specific initial velocity given, so we can include it as a constant of integration, \(C\).
Therefore,
\(v_x(t) = \frac{1}{2}\alpha t^2 + C\)
Next, to find the displacement function, we need to integrate the velocity function with respect to time:
\(x(t) = \int v_x(t) \, dt\)
Since \(v_x(t) = \frac{1}{2}\alpha t^2 + C\), we can substitute it into the equation:
\(x(t) = \int \left(\frac{1}{2}\alpha t^2 + C\right) \, dt\)
To integrate \(\frac{1}{2}\alpha t^2\) with respect to \(dt\), we use the power rule of integration again:
\(\int \frac{1}{2}\alpha t^2 \, dt = \frac{1}{6}\alpha t^3 + C_1\)
Here, \(C_1\) represents the constant of integration for the displacement function.
So, the displacement function becomes:
\(x(t) = \frac{1}{6}\alpha t^3 + C_1 t + D\)
where \(D\) is an integration constant.