Question

The acceleration, \(a\), of an object varies with time, \(t\), according to the formula \(a=t^{2}+3 t\). Given that velocity \(v\) is given by \(v=\int a(t) \mathrm{d} t\), find the velocity after 5 seconds given that the object is at rest at \(t=0\).

Short answer

Answer: The velocity after 5 seconds is 79.17 m/s.

Step-by-step solution

Find the integral of the acceleration function

In order to find the velocity function, we need to integrate the acceleration function. So we have: \(v(t) = \int a(t) dt = \int (t^2 + 3t) dt\)

Integrate

Now we'll integrate the function: \(v(t) = \int (t^2 + 3t) dt = \frac{1}{3}t^3 + \frac{3}{2}t^2 + C\), where \(C\) is the constant of integration.

Apply the initial condition

We know that the object is at rest at \(t=0\), which means the initial velocity is 0. So, we have: \(v(0) = 0 = \frac{1}{3}(0)^3 + \frac{3}{2}(0)^2 + C \Rightarrow C = 0\) This means the velocity function is: \(v(t) = \frac{1}{3}t^3 + \frac{3}{2}t^2\)

Find the velocity at \(t=5\)

Now we'll plug in \(t=5\) into the velocity function to find the velocity after 5 seconds: \(v(5) = \frac{1}{3}(5)^3 + \frac{3}{2}(5)^2 = \frac{1}{3}(125) + \frac{3}{2}(25) = 41.67 + 37.5 = 79.17\) So the velocity after 5 seconds is 79.17 m/s.

Key concepts

Integration

Integration is a fundamental concept in calculus that helps us find quantities like area, volume, and in this case, velocity from acceleration. When we integrate a function, we are essentially finding the 'antiderivative' or reverse of differentiation.
To integrate the acceleration function given by \(a = t^2 + 3t\), we calculate \(\int (t^2 + 3t) \, dt\).
This requires applying basic integration rules:

• The power rule: \(\int t^n \, dt = \frac{t^{n+1}}{n+1} + C\), where \(n\) is not -1.
• Constant coefficients remain the same.

Thus, integrating each term, we have:
\[v(t) = \int t^2 \, dt + \int 3t \, dt = \frac{1}{3}t^3 + \frac{3}{2}t^2 + C\]
where \(C\) is the constant of integration.

Velocity calculation

Calculating velocity from acceleration involves integrating the acceleration function. This process is crucial as it reveals how the speed of an object changes over time.
After integrating \(a = t^2 + 3t\), we get the velocity function:
\[v(t) = \frac{1}{3}t^3 + \frac{3}{2}t^2 + C\]
To find a specific velocity value, such as at \(t=5\) seconds, substitute \(t\) into the velocity equation:
\[v(5) = \frac{1}{3}(5)^3 + \frac{3}{2}(5)^2\]
Simplifying this:

• \(\frac{1}{3}(125) = 41.67\)
• \(\frac{3}{2}(25) = 37.5\)

Add these results to get 79.17 m/s.
This tells us the object's speed at 5 seconds.

Constant of integration

The constant of integration \(C\) emerges because integration can generate a family of functions. It accounts for any vertical shifts in these functions.
When solving problems like this one, initial conditions are used to solve for \(C\).
Here, we know the object is at rest at \(t=0\) (where velocity \(v(0) = 0\)), so we plug into the velocity function:
\[v(0) = \frac{1}{3}(0)^3 + \frac{3}{2}(0)^2 + C = 0\]
This simplifies to \(C = 0\), leaving us with a specific velocity equation:

• \(v(t) = \frac{1}{3}t^3 + \frac{3}{2}t^2\)

The constant helps complete the picture of the velocity function, ensuring it fits all known parameters.