Kinematics Problems

Fundamentals of 1D kinematics – Solution Procedure – Note on integration limits – Further example problems – Kinematics in 2D and 3D

Fundamentals of 1D kinematics

For motion along a one-dimensional path (straight or curved), define the following:

\(s\) is a coordinate axis along the path
\(\displaystyle v = \frac{ds}{dt}\) is the velocity along the path
\(\displaystyle a = \frac{dv}{dt}\) is the acceleration along the path, also known as the tangential acceleration

These can be written as differential relations:

\(ds = v \; dt\)
\(dv = a \; dt\)
\(a \; ds = v \; dv\) (This is found from the other two by eliminating \(dt\).)

With these equations one can solve many kinematics problems.

Solution Procedure

Identify, in terms of \(a\), \(v\), \(s\) and \(t\), what is given and what you need to find. At least one of these will be a function, such as \(a(s)\) or \(s(t)\).
Determine which function needs to be converted to which other function in order to find what you need. Complete the sentence “Given __ find __.”
Identify the “boundary conditions”, if any. You may be given the position or velocity at the initial time or final time. It could be that \(v=0\) at some location or time.
With this, determine which of the three differential relations contains all of the symbols you have identified. If this is not possible, consider going through the procedure twice. For example, if the problem says “given \(a(s)\) find \(s(t)\)”, you can solve “given \(a(s)\) find \(v(s)\)”, then solve “given \(v(s)\) find \(s(t)\)”.
Rearrange your differential relation such that, once you substitute in the given function (from step 1), you have only two variables in your equation. Next is the separation of variables: get one variable entirely on the left and the other on the right side of the equation. If you have a differential quantity on each side of the equation, draw the integral sign on each with appropriate limits. See more about this technique.
Now you just have a math problem to solve.

Example: Flight of the bumblebee

As a bee flies along a curvy path, its distance along that path is given by \(s = (2t^2 - 8t)\) m, where \(t\) is in seconds. Determine the time when the bee’s velocity is zero.

Following the solution procedure:

We are given \(s(t)\), the bee’s position as a function of time. We need to find a time \(t_1\) such that \(v(t_1) = 0\).
Given \(s(t)\) find \(v(t)\).
We are not given any boundary conditions. We won’t need them for this problem.
Only “\(ds = v \; dt\)” has \(s\) and \(v\) and \(t\).
Since we are given \(s(t)\) and need to find \(v(t)\), we can do it with \(\displaystyle v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 - 8t)\).
Differentiate to obtain \(\displaystyle v(t) = 4t - 8\), which is zero at time \(t_1 =2\) s.

Example: Ballistics test

A bullet is fired into a clay block with speed \(v_0\). While in the clay it decelerates as \(a = -kv^3\), where \(k\) is a constant. What is the bullet’s velocity as a function of time after entry?

Following the solution procedure:

We are given \(a(v)\). We need to find \(v(t)\).
Given \(a(v)\) find \(v(t)\).
For boundary conditions we have \(v=v_0\) when \(t=0\). If we define \(s\) as the distance into the clay, then \(s=0\) when \(t=0\).
With interest in \(a\), \(v\) and \(t\), we need \(dv = a \; dt\).
To separate the \(t\) and \(v\) variables, we should write that as \(\displaystyle dt = \frac{dv}{a} = - \frac{dv}{k v^3}\).
The math problem to solve is \(\displaystyle \int_0^t dt = -\frac{1}{k} \int_{v_0}^v v^{-3} \; dv\).

Integrate to complete the solution:

\[t = -\frac{1}{k} \left [ \left ( \frac{1}{-2} \right ) v^{-2} \right ]_{v_0}^v = \frac{1}{2k} \left ( v^{-2} - v_0^{-2} \right ) \]

So

\[ v(t) = \frac{1}{\sqrt{2 kt + v_0^{-2}}} \]

Note on integration limits

1. Meaning of limits

Integration limits specify the initial and final conditions. For example you may evaluate

\[\int_{t_0}^{t_1} dt = \left [ t \right ]_{t_0}^{t_1} = t_1 - t_0\]

to obtain “final minus initial” \(t\), usually written as \(\Delta t\).

It is common to leave the upper limit as a variable, such as

\[\int^s_0 ds = \int_{t_0}^{t} v(t)\; dt\]

This will give you \(s\) on the left and stuff that depends on \(t\) on the right. That is, you will obtain \(s(t)\), position as a function of time.

2. Indefinite integrals

For an integral to be physically meaningful it must have limits. For example, if you solve an indefinite integral such as the following

\[ \int v\; dv = \tfrac{1}{2} v^2 + C\]

you are really writing \(C\) as a substitute for the lower limit, and writing the upper limit as variable \(v\):

\[\int_{v_0}^{v}v\; dv = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2\]

So \(C = - \tfrac{1}{2} v_0^2\).

See the example problem below (accelerating bobsled) for an illustration of the different approaches.

3. Limit agreement

When applying limits, make sure they agree on both sides of the equation. For example, integrating \(v \; dt = ds\) gives

\[\int_{t_0}^{t} v(t)\; dt = \int_{s_0}^{s} ds\]

This implies that

at time \(t_0\), position is \(s_0\) (lower limits agree)

at time \(t\), position is \(s\) (upper limits agree)

Example: Accelerating bobsled

A bobsled gets a push and then accelerates down a curvy, icy ramp. Starting at \(1\) m/s, it accelerates along the track at a rate given by \(a = (2t - 6)\) m/s², where \(t\) is in seconds. What is the bobsled’s velocity when \(t=5\) s?

Given \(a(t)\) find \(v(t)\).
\(\displaystyle \int_{t_0}^{t} a(t)\; dt = \int_{v_0}^{v} dv = v(t) - v_0\).
Boundary conditions: at \(t_0 = 0\) the velocity is \(v_0 = 1\). At \(t = 5\)s we want the velocity.

To illustrate use of limits, I will solve this three different ways.

Method 1: Using numerical limits

Let \(v_1\) be the velocity at time \(t = 5\).

\[\int_{0}^{5} (2t - 6) \; dt = \int_{1}^{v_1} dv\]

The left side gives \(\displaystyle \left [ t^2 - 6 t \right ]_0^5 = (25 - 30) - 0 = -5\).

The right side gives \(\displaystyle \left [ v \right ]_1^{v_1} = v_1 - 1\).

So the solution is \(v_1 = -5 + 1 = -4\) m/s.

Method 2: Variable upper limit

This method gives the same solution, but also gives us a formula for the sled’s velocity at all times.

The initial conditions provide the lower limits \(t_0 = 0\) and \(v_0 = 1\). The upper limits are left as variables:

\[\int_{0}^{t} (2t - 6) \; dt = \int_{1}^{v} dv\]

The left side gives \(\displaystyle \int_{0}^{t} (2t - 6) \; dt = t^2 - 6t\).

The right side gives \(\displaystyle \left [ v \right ]_1^{v} = v - 1\).

Setting these equal, we obtain a formula for the velocity at all times:

\[v(t) = t^2 - 6t +1\]

Plug in \(t=5\) for the requested solution: \(v(5) = 25 - 30 +1 = -4\) m/s.

Method 3: Indefinite integrals

Without limits, we can write

\[\int (2t - 6) \; dt = \int dv\]

Which may be solved as

\[t^2 - 6t + C_1 = v + C_2\]

The two arbitrary constants \(C_1\) and \(C_2\) may be combined into a single arbitrary constant \(C = C_1 - C_2\). So we have \(t^2 - 6t + C = v\).

Plug in the initial conditions (\(v = 1\) when \(t = 0\)) to obtain \(C = 1\). Substitute this back in to obtain a formula for velocity versus time:

\[v(t) = t^2 - 6t +1\]

Plug in \(t=5\) for the solution: \(v(5) = 25 - 30 +1 = -4\) m/s.

Note: Method 2 is generally preferred. It takes no more effort than Method 1, but provides a more complete solution. Method 3 makes the initial conditions hidden, which may lead to confusion.

Further example problems

1.: A particle travels along a straight line with a speed \(v = (0.5 t^3 - 8t)\) m/s, where \(t\) is in seconds. Determine the acceleration of the particle when \(t=2\) s.

Given \(v(t)\) find \(a(t)\).
\(dv = a \; dt\)

Acceleration is \(\displaystyle a(t) = \frac{dv}{dt} = 1.5 t^2 -8\).

Set time to \(t=2\) s and solve for \(a\):

\[a = -2 \; \rm{m/s}^2\]

2.: A particle travels along a curved path with a speed of \(v = (20 - 0.05s^2)\) m/s, where \(s\) is in meters. Determine the tangential acceleration of the particle when it is at distance at \(s=15\) m along the path.

Given \(v(s)\) find \(a(s)\).
\(a \; ds = v \; dv\)

Since we are only concerned with the tangential acceleration, we can use the kinematics of 1D motion.

Separation of variables gives \(\displaystyle a(s) = v \frac{dv}{ds}\).

We need the derivative \(\displaystyle \frac{d}{ds} (20 - 0.05s^2) = -0.1s\).

Then \(\displaystyle a(s) = v \frac{dv}{ds} = (20 - 0.05s^2)(-0.1s) = -2s +0.005s^3\).

So at \(s=15\) m,

\[a = -13.1 \, \rm{m/s}^2\]

3.: A particle travels along a curved path with a speed given by \(v = (4t - 3t^2)\) m/s, where \(t\) is in seconds. Determine the position of the particle (along the path) when \(t=4\) s, given that \(s = 0\) when \(t = 0\).

Given \(v(t)\) find \(s(t)\).
\(ds = v \; dt\)

Separation of variables gives \(\displaystyle \int_{t_0}^{t} v(t)\; dt = \int_{s_0}^{s} ds = s(t) - s_0\).

Initial conditions are \(t_0 = 0\) and \(s_0 = 0\). So

\[s(t) = \int_{0}^{t} (4t - 3t^2) \; dt = ½ \cdot 4 t^2 - ⅓ \cdot 3 t^3 = 2 t^2 - t^3 \]

At \(t=4\) s,

\[s = -32 \; \rm{m}\]

4.: A particle travels along a straight line with an acceleration of \(a = (10 - 0.2s)\) m/s², where \(s\) is measured in meters. Determine the velocity of the particle when \(s=10\) m if \(v=5\) m/s at \(s=0\).

Given \(a(s)\) find \(v(s)\).
\(a \; ds = v \; dv\)

Acceleration is given by \(\displaystyle \int_{s_0}^{s} a(s)\; ds = \int_{v_0}^{v} v \; dv = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2\)

Initial conditions are \(v_0 = 5\) m/s and \(s_0 = 0\). So

\[ \tfrac{1}{2} v^2(s) - \tfrac{1}{2} \cdot 25 = \int_{0}^{s} (10 - 0.2s) \; ds = 10s - ½ \cdot 0.2 s^2 = 10s - 0.1 s^2 \]

Simplify this to obtain \(v(s)\):

\[v(s) = \sqrt{ 20s - 0.2 s^2 + 25 }\]

At \(s=10\) m,

\[v = 14.3 \; \rm{m/s}\]

5.: The velocity of a particle traveling along a straight line is \(v = v_0 - ks\), where \(k\) is constant. If \(s=0\) when \(t=0\), determine the position of the particle as a function of time.

Given \(v(s)\) find \(s(t)\).
\(ds = v \; dt\)

Velocity depends on \(s\), so separate variables to obtain

\[\int_{t_0}^t dt = \int_{s_0}^{s} \frac{ds}{v(s)}\]

Displacement is now found by \(\displaystyle t-t_0 = \int_{s_0}^{s} \frac{ds}{v(s)}\).

Initial conditions are \(s=0\) at time \(t=0\). So

\[t = \int_{0}^{s} \frac{ds}{(v_0 - ks)} = \left[ -\frac{1}{k} \ln (v_0 - ks) \right]_0^s = -\frac{1}{k} \ln \frac{v_0-ks}{v_0}\]

Solve by exponentiating both sides: \(\displaystyle \frac{v_0-ks}{v_0} = e^{-kt}\), so

\[s(t) = \frac{v_0}{k} \left(1 - e^{-kt} \right)\]

6.: The acceleration of rocket travelling upward is given by \(a = (6 + 0.02s)\) m/s², where \(s\) is in meters. Determine the time needed for the rocket to reach an altitude of \(s=100\) m. Initially, \(v=0\) and \(s=0\) when \(t=0\).

This requires two steps: \(a(s) \rightarrow v(s)\), then \(v(s) \rightarrow s(t)\).

For a symbolic solution, define \(A=6\) m/s², \(B=0.02 \; \rm{s}^{-2}\).

For \(a(s) \rightarrow v(s)\):

\[ \int_{0}^{s} (A+Bs) \; ds = \tfrac{1}{2} (v^2 - 0^2)\]

\[ \left[ As + \tfrac{1}{2} Bs^2 \right]_0^s = \tfrac{1}{2} v^2\]

\[ v(s) = \sqrt{ 2As + \tfrac{1}{2} Bs^2 } \]

For \(v(s) \rightarrow s(t)\):

\[ t - 0 = \int_{0}^{s} { \frac{dv}{\sqrt{ 2As + \tfrac{1}{2} Bs^2 }} } \]

This integral has a messy solution.

Kinematics in 2D and 3D

The methods explained above also apply to any particle in 2D or 3D motion. Simply treat each coordinate direction as an independent 1D problem.

position vector (vector that points to the particle location): \(\vec{r} = x(t) \, \hat{\imath} + y(t) \, \hat{\jmath} + z(t) \, \hat{k}\)

The magnitude of this vector is \(\displaystyle r = \sqrt{ x^2 + y^2 + z^2}\), the particle’s distance from the origin. It is not the same as \(s\), the distance along its path of motion that we used in 1D motion.

velocity vector (how fast is the particle moving, and in which direction): \(\displaystyle \vec{v} = \frac{d \vec{r}}{dt} = v_x(t) \, \hat{\imath} + v_y(t) \, \hat{\jmath} + v_z(t) \, \hat{k}\)
where \(\displaystyle v_x = \frac{dx}{dt}\), \(\displaystyle v_y = \frac{dy}{dt}\), and \(\displaystyle v_z = \frac{dz}{dt}\).

The magnitude of this vector is \(\displaystyle v = \sqrt{ v_x^2 + v_y^2 + v_z^2}\), the particle’s speed. This is the same as the absolute value of the \(v\) that we used in 1D motion.

Vector \(\vec{v}\) always points tangent to the particle’s path. A unit vector in the direction of motion (the tangential direction) can be found: \(\hat{u}_t = \vec{v} /v\).

acceleration vector (how is velocity changing): \(\displaystyle \vec{a} = \frac{d \vec{v}}{dt} = a_x(t) \hat{\imath} + a_y(t) \hat{\jmath} + a_z(t) \hat{k}\)
where \(\displaystyle a_x = \frac{d v_x}{dt}\), \(\displaystyle a_y = \frac{d v_y}{dt}\), and \(\displaystyle a_z = \frac{d v_z}{dt}\).

This vector \(\vec{a}\) can also be written as \(\vec{a} = \vec{a}_t + \vec{a}_c\), where

tangential acceleration is \(\displaystyle \vec{a}_t = \frac{d v}{dt} \hat{u}_t\)

centripetal acceleration is \(\displaystyle \vec{a}_c = \frac{v^2}{r} \hat{n}\)

The magnitude of the tangential acceleration (\(\frac{dv}{dt}\)) is the same as the acceleration \(a\) we considered in 1D motion. For the centripetal acceleration, \(r\) is the radius of curvature and \(\hat{n}\) points normal to the path, toward the center of curvature. For a straight (non-curving) path, the radius of curvature \(r \rightarrow \infty\); then \(a_c = 0\).

Example: Uniform circular motion

An electron in a cyclotron moves counter-clockwise in a circle of radius \(R\), with a constant speed \(v_o\). If we define the constant \(\omega = v_o/R\), then its velocity is given by \(\vec{v} = v_o (- \hat{\imath} \sin \omega t + \hat{\jmath} \cos \omega t)\). At time \(t=0\) the particle is at coordinates \((R,0)\). Determine the electron’s position and acceleration vectors as a function of time.

Separate the problem into \(x\) motion and \(y\) motion:

\(v_x(t) = - v_o \sin \omega t\)
\(v_y(t) = v_o \cos \omega t\).

For the \(x\) motion we have the problem “given \(v(t)\) find \(x(t)\)”:

\(dx = v_x \; dt\)

\(dx = - v_o \sin \omega t \; dt\)

\(\displaystyle \int_R^x dx = \int_0^t -v_o \sin \omega t \; dt\)

\(\displaystyle x - R = \left [ \frac{v_o}{\omega} \cos \omega t \right ]_0^t = R \left [ \cos \omega t \right ]_0^t = R (\cos \omega t - 1)\)

\(x(t) = R \cos \omega t\)

We also have the problem “given \(v(t)\) find \(a(t)\)”:

\(d v_x = a_x \; dt\)

\(\displaystyle a_x = \frac{d v_x}{dt} = \frac{d}{dt} (- v_o \sin \omega t) = - v_o \omega \cos \omega t = - \frac{v_o^2}{R} \cos \omega t\)

\(\displaystyle a_x(t) = - \frac{v_o^2}{R} \cos \omega t\)

For the \(y\) motion we very similar problems. You may verify the solutions are

\(y(t) = R \sin \omega t\)

\(\displaystyle a_y(t) = - \frac{v_o^2}{R} \sin \omega t\)

The final solutions are

\(\vec{r}(t) = R (\hat{\imath} \cos \omega t + \hat{\jmath} \sin \omega t)\)

\(\displaystyle \vec{a}(t) = - \frac{v_o^2}{R} (\hat{\imath} \cos \omega t + \hat{\jmath} \sin \omega t)\)