Answer:
A. (I) v = 46.42 m/s; (ii) v = 47.35 m/s; (III) v = 48.09 m/s; (iv) v = 48.26 m/s; (v) v = 58.28 m/s
B. v = 48.28 m/s
Note: the question is missing some values. The full Question is provided below:
If an arrow is shot upward on Mars with a speed of 52 m/s, its height in meters t seconds later is given by y = 52t − 1.86t2. (Round your answers to two decimal places.)
(a) Find the average speed over the given time intervals. (i) [1, 2] m/s (ii) [1, 1.5] m/s (iii) [1, 1.1] m/s (iv) [1, 1.01] m/s (v) [1, 1.001] m/s
(b) Estimate the speed when t = 1. m/s
Step-by-step explanation:
Height, y = 52t - 1.86t²
Velocity = ∆y/∆t = 52 - 1.86 * 2t = 52- 3.72t
A. Average velocity = (v1 + v2)/2
(i) At t = 1, 2
Average velocity = (52 - 3.72*1 + 52 -3.72*2)/2 = 46.42 m/s
(ii) At t = 1,1.5
Average velocity = (52 - 3.72*1 + 52 - 3.72*1.5)/2 = 47.35 m/s
(iii) At t = 1,1.1
Average velocity = (52 - 3.72*1 + 52 -3.72*1.1)/2 = 48.09m/s
(iv) At to = 1, 1.01
Average velocity = (52 - 3.72*1 + 52 - 3.72*1.01)/2 = 48.26 m/s
(iv) At t = (1, 1.001)s
Average velocity = (52 - 3.72*1 + 52 - 3.72*1.001)/2 = 48.28 m/s
B. Speed at t = 1s
Velocity = 52 - 3.72 * 1 = 48.28 m/s
