Integrals of Motion

A particle is traveling with velocity \( v(t) = 6t^2, \) and has displacement \(100\) after 5 seconds. Find the velocity function of this particle.

Solution:

The first step is to take the indefinite integral of the particle.

\[ \begin{align} s(t) & = \int v(t) \, \mathrm{d}t \\ & = \int 6 t^2 \, \mathrm{d}t \\ & = 2 t^3 + c. \end{align} \]

Now, you can substitute \(t=5\) into the displacement vector to find the value of the constant of integration.

\[ \begin{align} v(5) = 100 & = 2 \cdot 5^3 + c \\ & = 2 \cdot 125 + c \\ & = 250 + c. \end{align} \]

Hence, you have that

\[ \begin{align} 250 + c & = 100 \\ \implies c & = -150. \end{align} \]

Hence, the displacement function must be

\[ s(t) = 2 t^3 - 150. \]

A particle is traveling with acceleration \( a(t) = \cos{t} \) and is stationary at \(t=0.\) Find the velocity function for this particle.

Solution:

First, integrate the acceleration function.

\[ \begin{align} v(t) & = \int a(t) \, \mathrm{d}t \\ & = \int \cos{t} \, \mathrm{d}t \\ & = \sin{t} + c. \end{align} \]

Now, since the particle is stationary at \(t=0,\) the velocity must be \(0.\) Substitute this in to find the value of \(c.\)

\[ v(0) = 0 = \sin(0) + c = c \]

Hence, \(c\) must be \(0.\) This means the final velocity function must be

\[ v(t) = \sin{t}. \]

A particle travels with velocity \(v(t) = 8 -4t.\) Find the total displacement and the total distance of the particle between times \(t=1\) and \(t=3.\)

Solution:

First, find the total displacement. Take the definite integral of the velocity function between \(t=1\) and \(t=3.\)

\[ \begin{align} \int_1^3 8 - 4t \, \mathrm{d}t & = [8t - 2t^2]_1^3 \\ & = (8 \cdot 3 - 2 \cdot 3^2) - (8 \cdot 1 - 2 \cdot 1^2) \\ & = (24 - 18) - (8 - 2) \\ & = 6 - 6 \\ & = 0. \end{align} \]

So the total displacement is \(0.\)

To find the total distance travelled, you must find the integral of the absolute value of the function.

\[ \int_1^3 | 8 - 4t | \, \mathrm{d}t. \]

The function will go negative at \(t=2,\) hence the integral can be split into two at this point.

\[ \begin{align} \int_1^3 | 8 - 4t | \, \mathrm{d}t & = \int_1^2 8 - 4t \, \mathrm{d}t + \int_1^2 -(8-4t) \, \mathrm{d}t \\ & = \int_2^3 8 - 4t \, \mathrm{d}t + \int_1^2 -8+4t \, \mathrm{d}t. \end{align} \]

Now you can evaluate the integrals on the left to deduce the total distance travelled.

\[ \begin{align} \int_1^3 | 8 - 4t | \, \mathrm{d}t & = [8t - 2t^2]_1^2 + [-8t + 2t^2]_2^3 \\ & = [(8\cdot 2 - 2\cdot 2^2) - (8 \cdot 1 - 2 \cdot 1^2)] + [(-8\cdot 3 + 2 \cdot 3^2) - (-8 \cdot 2 - 2 \cdot 2^2)] \\ & = [(16 - 8) + (8 - 2)] + [(-24 + 18) - (-16 + 8)] \\ & = [8 - 6] + [-6 + 8] \\ & = 4. \end{align} \]

So the total distance travelled is \(4.\)

A particle travels with velocity \(v_1(t) = t^2 \) for 3 seconds, and then has velocity \(v_2(t) = 3t\) for a further 6 seconds. Calculate the total distance travelled by the particle in this time.

Solution:

First, since the question asks for the total distance travelled, you must take the absolute values of the velocities. But since both of the velocities are always positive, the total distance travelled will be equal to the displacement, and the absolute values are irrelevant. Hence, the total distance travelled is

\[ \begin{align} \int_0^3 3t^2 \, \textrm{d}t + \int_3^9 3t \, \textrm{d}t & = [t^3]_0^3 + \left[\frac{3}{2} t^2 \right]_3^9 \\ & = (3^3 - 0^3) + \left(\frac{3}{2} \cdot 9^2 - \frac{3}{2} \cdot 3^2 \right) \\ & = 27 + (\frac{243}{2} - \frac{27}{2}) \\ & = 27 + 116 \\ & = 143 \end{align} \]

So the total displacement and total distance travelled are 143 units.

The acceleration of a particle is given by \[a(t) = 6t\] and the position of the particle is \(10\) at \(t=0\) and \(14\) at \(t=2.\) calculate the displacement function of the particle.

Solution:

This time, you will have to integrate twice, since the function given is acceleration and not velocity. Integrating the function once will give you:

\[ \begin{align} v(t) & = \int a(t) \, \textrm{d}t \\ & = \int 6t \, \textrm{d}t \\ & = 3t^2 + c. \end{align}\]

Now, you can integrate the velocity function to find the displacement function.

\[ \begin{align} s(t) & = \int v(t) \, \textrm{d}t \\ & = \int 3 t^2 + c \, \textrm{d}t \\ & = t^2 + ct + d. \end{align} \]

\(d\) is another constant of integration. Now, you can substitute in the initial conditions to find the exact displacement function. If \(t=0:\)

\[ \begin{align} s(0) = 10 & = 0^2 + c \cdot 0 + d \\ \implies d & = 10. \end{align} \]

Hence, \(d\) must be \(0.\) If \(t = 2:\)

\[ \begin{align} s(2) = 14 & = 2^2 + c \cdot 10 + 10 \\ & = 14 + 10c \\ \implies 0 & = 10c \end{align} \]

so \(c = 0.\) Hence, the final formula for displacement is

\[ s(t) = t^2 + 10. \]

Particle \(A\) is travelling at velocity \(v_A(t) = 2t^2 \) and particle \(B\) is travelling at velocity \(v_B(t) = 4t.\) Both particles are in the same position at time \(t=0.\)

1. Find the distance between the particles at time \(t=5.\)
2. At what time are the particles in the same position again?

Solution:

1) Since the particles are only moving in one dimension, the distance between them will simply be the absolute value of the difference between their distances from the starting points.

\[ \begin{align} \text{Distance between A and B at time 5} & = \int_0^5 v_A(t) \, \textrm{d}t - \int v_B(t) \, \textrm{d}t \\ & = \int v_A(t) - v_B(t) \, \textrm{d}t. \end{align} \]

Now, you can plug the formulas for \(v_A(t)\) and \(v_B(t)\) into the formula above to get the distance between them.

\[ \begin{align} \int_0^5 v_A(t) - v_B(t) \, \textrm{d}t & = \int_0^5 2t^2 - 4t \, \textrm{d}t \\ & = \left[ \frac{2}{3} t^3 - 2 t^2 \right]_0^5 \\ & = \frac{2}{3} \cdot 5^3 - 2 \cdot 5^2 \\ & = \frac{250}{3} - 50 \\ & = \frac{100}{3}. \end{align} \]

So the distance between the two particles at time \(t=0\) is \(\frac{100}{3}.\)

2) This problem can be solved in the same way as question one, but you must use a variable, call it \(t',\) as the upper bound of integration, then set the result as equal to \(0\) and solve the equation for \(t'.\) Using the same steps as before, you will get that:

\[ \begin{align} \int_0^5 v_A(t) - v_B(t) \, \textrm{d}t & = \int_0^5 2t^2 - 4t \, \textrm{d}t \\ & = \left[ \frac{2}{3} t^3 - 2 t^2 \right]_0^{t'} \\ & = \frac{2}{3} t'^3 - 2 t'^2. \end{align} \]

Now you can set this equal to 0, and solve for \(t'.\)

\[ \begin{align} \frac{2}{3} t'^3 - 2 t'^2 & = 0 \\ & \implies t' (\frac{2}{3} t' - 2 ) = 0. \end{align} \]

For this equation to be true, either \(t'=0\) or \(\frac{2}{3} t' - 2 = 0.\) \(t'=0\) is simply the starting point, and this is not the answer that the question is looking for. Hence, it must be that

\[ \begin{align} \frac{2}{3} t' - 2 & = 0 \\ \implies t' & = 2 \cdot \frac{3}{2} \\ \implies t' & = 3\ \end{align} \]

Hence, the particles will be in the same position again at \(t=3.\)