QUIZ #1: CALCULUS 1A (Stankova) Wednesday, January 28, 2004 Section 10:00–11:00 (Voight)

Problem 1. The position of a hydrogen fuel-cell vehicle is given by the values in the following table:

t (seconds) 0 5 10 15 20  25 ; s (feet)  30 150 450 950 1600 2575

(a) Find the average velocity for the time period beginning when t = 5 and lasting: (i) 15 seconds (ii) 10 seconds (iii) 5 seconds

(b) Use the graph of s as a function of t to estimate the instantaneous velocity when t = 5.

Solution. The average velocity is given by the distance traveled divided by the amount of time. A time period beginning when t = 5 and lasting 15 seconds ends at t = 20 seconds. Therefore for (i), the average velocity over [5, 20] is

v = \frac{s(20) − s(5)}{20 − 5} = \frac{1600 − 150}{15} = \frac{1450}{15} = \frac{290}{3} ft/s.

[You could have left the fraction unsimplified!] Similarly, for (ii) we have

\frac{950 − 150}{15 − 5} = \frac{800}{10} = 80 ft/s

and for (iii) we have

\frac{450 − 150}{10 − 5} = \frac{300}{5} = 60 ft/s.

For (b), we graph with the proper scale the points given in the graph as follows:

We fit a curve to the points (0, 30), (5, 150), …, (25, 2575).

The tangent line appears to go through the point (15, 500), so we estimate its slope to be

\frac{500 − 150}{15 − 5} = 35 ft/s.

The instantaneous velocity is exactly this slope. [It is acceptable to estimate the slope in other ways using the graph, as long as you justify your answer.] QUIZ #1: CALCULUS 1A (Stankova) Wednesday, January 28, 2004 Section 11:00–12:00 (Voight)

Problem 1. If a pellet is shot upward by the Mars rover, its height in kilometers after t seconds is given by h = 3t − t^{2}.

(a) Find the average velocity over the given time intervals: (i) [0, 1] (ii) [0, 0.5] (iii) [0, 0.1]

(b) Graph the height of the pellet as a function of time; draw the tangent line to the graph at time t = 0.

(c) Find the initial velocity at which the pellet was shot.

Solution. The average velocity is given by the distance traveled divided by the amount of time. Therefore the average velocity for (i) is

v = \frac{h(1) − h(0)}{1 − 0} = (3(1) − 1^{2}) − 0 = 2 m/s.

Similarly, for (ii) we have

v = \frac{3(0.5) − (0.5)^{2} − 0}{0.5 − 0} = \frac{1.5 − 0.25}{.5} = 2(1.25) = 2.5 m/s

and for (iii) we have

v = \frac{3(0.1) − (0.1)^{2} − 0}{0.1 − 0} = \frac{0.3 − 0.01}{.1} = \frac{0.29}{.1} = 2.9 m/s.

For (b), we plot points: we find that at t = 0 the height is h = 0, at t = 1 the height is h = 2, and so on:

For (c), notice that the slope of the tangent line, which is the same as the initial velocity or the instantaneous velocity at t = 0, is approximately 3 m/s. Also, the average velocities computed in (a) (none other than the slopes of the secant lines) approach this same value, 3 m/s.