Calculate How Fast an Object Was Moving at a Particular Time

Question

The question is based from a table (posted below):

Approximately how fast was the ball traveling when $t = 1.0$?

I calculated the average speed of the ball during the first second ($16$ ft/sec), the average speed during the half-second interval (from $t = 0.5$ to $t = 1.0$) which was $24$ ft/sec, the average speed from $t = 0.9$ to $t = 1.0$ ($30$ ft/sec) and from $t = 1.0$ to $t = 1.1$ (which was $34$ ft/sec). However, I do not know how to use that information to solve for the individual speed of the ball at $t = 1.0$. Any help will be appreciated!

Answer 1

As mentioned in comments, the statement of the equation "Approximately how fast was the ball moving when $t = 1.0~\text{s}$?" suggests that we are interested in its instantaneous speed rather than its instantaneous velocity at $t = 1.0~\text{s}$ since no direction is specified.

The table provides the height of the ball at time intervals of $0.1~\text{s}$. What we want to approximate is its instantaneous speed at $t = 1.0~\text{s}$.

The average speed of the ball over a time interval $t$ is $$v_{av} = \frac{|\Delta h|}{\Delta t}$$ where $\Delta h$ is the change in its height and $\Delta t$ is the length of the time interval during which that change occurs. The instantaneous speed is $$v = \lim_{\Delta t \to 0} \frac{|\Delta h|}{\Delta t}$$ Since we only know the height of the object at discrete time intervals, we can best approximate how fast the ball is moving by finding how far the ball moves in the smallest time interval that includes $t = 1.0~\text{s}$.

There are two such intervals, $[0.9, 1.0]$ and $[1.0, 1.1]$. During the first interval, the ball falls $(67.0 - 64.0)~\text{ft} = 3.0~\text{ft}$. During the second time interval, the ball falls $(64.0 - 60.6)~\text{ft} = 3.4~\text{ft}$. Averaging these two results gives $$v_{av} = \frac{3.0 + 3.4}{2 \cdot 0.1}~\frac{\text{ft}}{\text{s}} = \frac{6.4}{0.2}~\frac{\text{ft}}{\text{s}} = 32~\frac{\text{ft}}{\text{s}}$$ Equivalently, we could find the average velocity over the time interval $[0.9, 1.1]$.

Not coincidentally, this happens to be how fast we would expect the ball to be moving in free ball in the Earth's gravitational field. If an object is in free ball, its speed $t~\text{s}$ after it begins falling is $$v = gt$$ where $g = 32~\text{ft}/\text{s}$ (in English units).

In the comments, you asked about finding the approximate speed at $t = 1.8~\text{s}$. If we find the average over the time interval $[1.7, 1.9]$, we obtain $$v_{av} = \frac{33.8 - 22.2}{1.8 - 1.6}~\frac{\text{ft}}{\text{s}} = \frac{11.6}{0.2}~\frac{\text{ft}}{\text{s}} = 58~\frac{\text{ft}}{\text{s}}$$ Applying the formula $v = gt$ when $t = 1.8~\text{s}$ gives $$v = 32~\frac{\text{ft}}{\text{s}} \cdot 1.8~\text{s} = 57.6~\frac{\text{ft}}{\text{s}} \approx 58~\frac{\text{ft}}{\text{s}}$$ We cannot directly compute the speed of the ball $2.2~\text{s}$ after it is released. If we assume that we are dealing with a ball in free fall (no air resistance, no initial velocity), then we could use the formula $v = gt$ to approximate its speed.

Answer 2

You can fit it to a local curve. Let $h(t) = a+bt+ct^2$ where $a, b, c$ are constants. Use the set of data points to find the constants $t=(.9, 1, 1.1)$ and $h=(67,64,60.6)$ by evaluating $h(t)$ at each point and then solve the non-linear system for $a,b,c$. At this point, you can take the derivative of $h$ which is $\dfrac{dh}{dt}(t) = b + 2ct$ and evaluate at $t=1$ to get the velocity.

Answer 3

$$h(t)=-16t^2 + v_0t + 80$$ $$h(1) = 64 = -16+v_0+80=64+v_0\implies v_0=0$$

Therefore, $$h(t) = -16t^2+80$$

Taking the derivative gives $$\dot h(t)=-32t$$

Finally, $$\dot h(1) = -32$$

so the object is moving at 32 feet per second downward.