Physics 100
West Chester University
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Lesson 4
1D Kinematics - Acceleration and Falling Motion

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Acceleration
We define acceleration as the change in velocity over the change in time:
a = {\Delta v \over \Delta t}

As with velocity, acceleration is a vector quantity. Thus we can write:
a_x = {\Delta v_x \over \Delta t}

and
a_y = {\Delta v_y \over \Delta t}

Example

Given that a runner starts from 0 m/s and accelerates to 10 m/s over a period of 5 seconds, what is the runner's acceleration?

a_x = {(10m/s - 0m/s) \over (5 s)}

a_x = {10 m/s \over 5 s} = 2 m/s^2

Example

Given that a ball falls at -9.8 m/s2, when will it's velocity equal -39.2 m/s2?

a_y = {v_{yf} - v_{yi} \over \Delta t}

\Delta t = {v_{yf} - v_{yi} \over a_y}

\Delta t = {-39.2 m/s - 0 m/s \over -9.8 m/s^2}

\Delta t = 4 s

As with the distance equations, one can manipulate the acceleration equation to find different information (time, velocity or acceleration). But to this point, we have not considered constant acceleration and how it will affect distance calculations. First let us consider the example of falling motion.

t
v
 \bar v
\Delta d = \bar {v \over {\Delta t}}
 \Delta d_{tot}
0 s
 0 m/s
  -
-0 m
 -0 m
1 s
 -10 m/s
 -5 m/s
 -5 m
 -5 m
2 s
 -20 m/s
 -15 m/s
 -15 m
 -20 m
3 s
 -30 m/s
 -25 m/s
 -25 m
 -45 m
4 s
 -40 m/s
 -35 m/s
 -35 m
 -80 m
If we wanted to get the results of this table using a formula, we could use the following:

\Delta y = v_{iy}\Delta t + {1 \over 2}a_y \Delta t

\Delta y = (0 m/s) * (4 s) + {1 \over 2} (-10 m/s^2) * (4 s)^2

\Delta y = (-5 m/s^2) * (16 s^2)

\Delta y = -80 m

The above formula is a shorthand for the table, which ultimately is a work-saver in many cases. To appreciate this, let's do a problem where there is a non-zero initial velocity. Presume that we launch a ball upwards at 30 m/s and we want to know how high it goes. First, we use the understanding that when the ball reaches the apex of flight, it's velocity, vfy, will be zero. Then using our formula for acceleration, velocity, and time, we get:

\Delta v = v_{fy} - v_{iy} = a_y \Delta t

\Delta t = {v_{fy} - v_{iy} \over a_y}

\Delta t = {0 m/s - 30 m/s \over -10 m/s^2}

Or
\Delta t = 3 s

is the time it takes to get to the apex. We can now use this to determine the height of the apex:
\Delta y = v_{iy} \Delta t + {1 \over 2} a_y (\Delta t)^2

Plugging in numbers, we get the following height for the apex:
\Delta y = (30 m/s)(3 s) + {1 \over 2} (-10 m/s^2) (3 s)^2

\Delta y = 90 m + (-5 m/s^2)(9 s^2)

\Delta y = 45 m

It turns out that there is another way to do this problem that is easier. Unfortunately, we need to derive a formula to procede. Earlier on, we showed that:
1) \Delta y = \bar v_y \Delta t

The quantity \bar v_y is the average velocity. The change in time \Delta t may be represented in a different manner as well.
2) \Delta t = {\Delta v \over a_y}

The average velocity is:
3) \bar v_y = {1 \over 2} (v_{fy} + v_{iy})

Substitution of this formula into equation 1) gives:
4) \Delta y = {1 \over 2} (v_{fy} + v_{iy})\Delta t

Substitution of equation 2) into 4) gives:
5) \Delta y = {1 \over 2a_y} (v_{fy} + v_{iy})(v_{fy} - v_{iy})

Finally, by FOILing the velocity terms in the numerator, we get an expression for \Delta y:
\Delta y = {v_{fy}^2 - v_{iy}^2 \over 2a_y}

Now, plugging in :
\Delta y = {0 m^2/s^2 - (30 m/s)^2 \over 2(-10 m/s^2)}

\Delta y = {-900 m^2/s^2 \over -20 m/s^2} = 45 m

This is the same as what we have from above, but without all the extra work (aside from the derivation). Remembering the results of our derivation, it seems, will be useful in saving work.