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.

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

c

step-by-step explanation: