15 Modeling Motion

Kinematics

Based on the definition of acceleration as the rate of change of the velocity we can calculate the change in velocity during a time interval as the acceleration multiplied by the length of the time interval:

(1)   \begin{equation*} \bold{\Delta v} = \bold{a}\Delta t \end{equation*}

Reinforcement Exercises

If a person has an acceleration of 5.0 m/s/s, how much does their velocity change in 2.0 s?

We can find the current velocity by adding the expression for change in velocity to the initial velocity:

(2)   \begin{equation*} \bold{v_f} = \bold{v_i} + \bold{a}\Delta t \end{equation*}

Reinforcement Exercises

If  the person in the previous exercise has an initial velocity of  2.0 m/s, what is their new velocity after the 2.0 s?

We can calculate the average velocity during the interval as the average of the initial and final velocities:

(3)   \begin{equation*} \bold{v_{ave}} = \frac{\bold{v_i} + \bold{v_f}}{2} \end{equation*}

Reinforcement Exercises

What is the average velocity of the person in the previous exercise?

Using the definition of velocity as the rate of change of position we can calculate the change in position during a time interval as the average velocity during the interval multiplied by the length of the time interval.

(4)   \begin{equation*} \bold{\Delta x} = \bold{v_{ave}}\Delta t \end{equation*}

Reinforcement Exercises

What is the change in position of the person in the previous exercises?

Adding the above expression for change in position to the initial position allows us to calculate the final position after any time:

(5)   \begin{equation*} \bold{x_f} = \bold{x_i} +\bold{v_{ave}}\Delta t \end{equation*}

Reinforcement Exercises

If  the person in the previous exercises started at a position of 4 m/s, what is their final position?

Reinforcement Activities

Some algebra allows us to combine everything the from previous steps into a single equation that can save some time on some problems. the result looks like this:

(6)   \begin{equation*} \bold{x_f} = \bold{x_i} + \bold{v_i}\Delta t + \frac{1}{2}\bold{a}(\Delta t)^2 \end{equation*}

If you want to know what that algebra looks like, here it is:
To get the above equation we used equation (3) to replace the average velocity with the expression for average velocity:

(7)   \begin{equation*} \bold{x_f} = \bold{x_i} + \frac{\bold{v_i} + \bold{v_f}}{2}\Delta t \end{equation*}

Using equation (2) we can then replace the final velocity:

(8)   \begin{equation*} \bold{x_f} = \bold{x_i} + \frac{(\bold{v_i} + \bold{v_i}+ \bold{a}\Delta t)}{2}\Delta t \end{equation*}

After some simplification we are there:

(9)   \begin{equation*} \bold{x_f} = \bold{x_i} + \bold{v_i}\Delta t + \frac{1}{2}\bold{a}(\Delta t)^2 \end{equation*}

Everyday Example

After leaving a friend’s 3rd story apartment you get to your car and realize that you have left your keys in the apartment. You call your friend and ask them to drop the keys out the window to you. We want to figure out how long it will take the keys to reach you and how fast they will be falling when they get there. The third story window is about 35 ft off the ground. We can convert to meters and use our previously stated acceleration for falling objects, g =9.8 m/s/s, or we can stick with feet and use g = 32 ft/s/s, so let’s do that.

Starting from our last equation from the work we did above:

    \begin{equation*} \bold{x_f} = \bold{x_i}+ \bold{v_i}\Delta t + \frac{1}{2}\bold{a}(\Delta t)^2 \end{equation*}

We choose upward as our positive direction and the ground as our origin, therefore our initial position is 35 ft and our final position is 0 ft. The keys are dropped from rest, so our initial velocity is zero. Putting the zeros into the equation above we have:

    \begin{equation*} 0= \bold{x_i} + 0 + \frac{1}{2}\bold{a}(\Delta t)^2 \end{equation*}

Now we can isolate the time variable:

(10)   \begin{equation*} t^2 = \frac{-2\bold{x_i}}{\bold{a}} \end{equation*}

Take the square root to find the time

(11)   \begin{equation*} t = \sqrt{\frac{-2\bold{x_i}}{\bold{a}}} \end{equation*}

Entering our known values we can find the fall time. We will use -32 ft/s/s for our acceleration because the acceleration due to gravity is downward and we have chosen upward as the positive direction.

(12)   \begin{equation*} t = \sqrt{\frac{-2(35\,\bold{ft})}{-32\,\bold{ft/s/s}}} = 1.479 \,\bold{s} = 1.5 \,\bold{s} \end{equation*}

Lastly, we can find the velocity of the keys using equation (2) above

(13)   \begin{equation*} \bold{v_f} = \bold{v_i} + \bold{a}\Delta t = 0 + (32\,\bold{ft/s/s})(1.479\,\bold{s}) = 47 \,\bold{ft/s} \end{equation*}

The final velocity of 47 ft/s is about 32 MPH. If the keys smack your hand at that speed, it will hurt. There are techniques you could use to prevent injury in such a situation, and those techniques will be the topic of the next Unit.

In solving the previous example we found an equation to calculate the time required for an object with a certain acceleration to reach a final position of zero when starting from a known initial position. Among other things, this allows us to calculate the time required to fall to the ground from a certain starting height. That equation will come up often, so lets write it out here:

(14)   \begin{equation*} t = \sqrt{\frac{-2\bold{x_i}}{\bold{a}}} \end{equation*}

If acceleration is set to  -9.8 m/s/s (or -g), then this equation calculates the free-fall time for a choice of negative as the downward direction.

Reinforcement Exercises

 

The Jerk

We have learned in the last few chapters that our example skydiver has an initial acceleration of 9.8 m/s/s and an acceleration of zero after reaching terminal velocity, so between those points the acceleration must be changing.  The rate of change of the acceleration is known as the jerk, but we won’t deal with jerk in this textbook and will instead focus on motion with constant acceleration. However, if we really want to analyze our skydiver’s full motion, we will need to somehow deal with a changing acceleration. That’s what the ne

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