A model rocket leaves the ground at time t=0 and travels straight up from the ground.

Question

The height, in feet, of the rocket above the ground is given by y(t), where t is measured in seconds for 0≤t≤120. Values of y(t) for selected values of t are given in the table above. Of the following values of t, at which value would the velocity of the rocket most likely be greatest based on the data in the table?

t (seconds) 0 20 40 60 80 100 120

y(t) (feet) 0 105 300 900 2400 5000 10000

A) t = 20

B) t = 40

C) t = 60

D) t = 80

Answer 1

If you graph this, the answer will be obvious.

The velocity is the change in height over the change in time, aka, the slope of the height vs time graph. The slope gets larger and larger as t gets bigger so t = 120 has the highest slope. To prove this by the data, calculate the slope (y₂ - y₁)/(x₂ - x₁) for two adjacent points. For instance, the slope using points (100, 5000) and (120, 10000) yield a slope of 5000/20 = 250 ft/s while the slope using points (0, 0) and (20, 105) yield a slope of 105/20 = 5.25 ft/s.

It can be tricky using two adjacent points to calculate the slope because you don't know whether to use the point before t = 40 or the point after t = 40 to calculate the slope at t = 40. The best results would be to average the two. But, for this exercise, just be consistent so you can compare.