The table below gives the height, h, of a falling object (in meters) at various times, t (in seconds).

Question

The table below gives the height, h, of a falling object (in meters) at various times, t (in seconds).

h (m)	47.69	47.01	46.24	45.39	44.47	43.46	42.38
t (seconds)	0.7	0.8	0.9	1.0	1.1	1.2	1.3

Compute the average velocity of the object on each of the following intervals. Use two decimal places of accuracy for each average velocity.

1. [0.7, 1.0] 
2. [0.8, 1.0] 
3. [0.9, 1.0] 
4. [1.0, 1.1] 
5. [1.0, 1.2] 
6. [1.0, 1.3] 
7. Estimate the instantaneous velocity when t = 1 s. Be accurate to at least one decimal place. 

Answer 1

To find an average velocity between two points of time from a table, apply:

V_{avg = Δs/Δt}

With Δs being the change in displacement of an object and Δt being change in time. Keep in mind to calculate these values, you would apply (final value - initial value)

So if we wanted lets say we wanted the average velocity from 0.7 to 1.0. Or theoretically 0.7 seconds to one second, we would calculate:

V_{avg = (45.39 - 47.69) / ( 1.0 - 0.7) OR about -7.67 m/s.}

_{The same would be applied for every subsequent interval from the table. You are essentially just finding the velocity or rate of change from position and time of an object from one time point to another. In that span of 0.3 seconds, the object goes at a speed of -7.67 m/s as it is falling down.}

NOTE: This is only the average rate of change. To find an instantaneous rate of change we MUST use the slope of the secant line through the two closest values to a target value.

So for t = 1s, we would use t = 0.9 seconds and t = 1.1 seconds to find the slope of our secant line.

h(1.1) - h(0.9) / (1.1-0.9) = (44.47 - 46.24) / (0.2) = -8.85 m/s.

This means at the INSTANCE or exact moment of 1 second, the object is going -8.9 m/s, Later on, you may use a derivative to calculate this value.

Hope this helps!