Math 1271 - Solutions to Quiz 1
1. If a ball is thrown in the air with a velocity of 40 ft/s, its height in feet t seconds later is
given by y = 40t − 16t2 .
(a)(8 points) Find the average velocity for the time period beginning when t = 2 and lasting:
(i) t = .5 seconds, (ii) t = .1 seconds, (iii) t = .05 seconds, (iv) t = .01 seconds
ANSWER:
(i) The average velocity is given by the slope of the secant line. Therefore, we are trying to
find the slope of the secant line between the points that correspond to t = 2, and t = 2.5
(which is .5 seconds later). Letting f (t) = 40t − 16t2 , we want to find:
[40(2) − 16(2)2 ] − [40(2.5) − 16(2.5)2 ]
16 − 0
f (2) − f (2.5)
=
=
= −32f t/s
2 − 2.5
2 − 2.5
−.5
(ii) Similar to above:
f (2) − f (2.1)
[40(2) − 16(2)2 ] − [40(2.1) − 16(2.1)2 ]
16 − 13.44
=
=
= −25.6f t/s
2 − 2.1
2 − 2.1
−.1
(iii) Again by a similar process, we get −24.8f t/s
(iv) And −24.16f t/s
(b)(2 points) Estimate the instantaneous velocity when t = 2.
ANSWER:
The values above are approaching −24f t/s as the time intervals get smaller (in other words,
as we get closer to t = 2), and so we estimate our instantaneous velocity when t = 2 to be
−24f t/s.