Question

If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by y = 56t − 1.86t2.
(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.

Answer 1

Answer: (a) (i)50.42 m/s (ii) 51.35 m/s (iii) 52.094 m/s (iv) 52.2614 m/s (v) 52.27814 m/s

(b) 54.14 m/s

Step-by-step explanation:

The average speed over the given time interval [a,b]: = $\frac{y(b)-y(a)}{b-a}$

Given: If an arrow is shot upward on Mars with a speed of 56 m/s, its height in meters t seconds later is given by $y = 56t − 1.86t^2.$

(a)

(i) average speed = $\frac{y(2)-y(1)}{2-1}$

$=\frac{56(2)-1.86(2)^2-(56(1)-1.86(1)^2)}{1}$

$=50.42$ m/s

(ii) average speed = $\frac{y(1.5)-y(1)}{1.5-1}$

$=\frac{56(1.5)-1.86(1.5)^2-(56(1)-1.86(1)^2)}{0.5}$

$=51.35$m/s

(iii) average speed = $\frac{y(1.1)-y(1)}{1.1-1}$

$=\frac{56(1.1)-1.86(1.1)^2-(56(1)-1.86(1)^2)}{0.1}$

$=52.094$m/s

(iv) average speed = $\frac{y(1.01)-y(1)}{1.01-01}$

$=\frac{56(1.01)-1.86(1.01)^2-(56(1)-1.86(1)^2)}{0.01}$

$=52.2614$m/s

(v) average speed = $\frac{y(1.001)-y(1)}{1.001-01}$

$=\frac{56(1.001)-1.86(1.001)^2-(56(1)-1.86(1)^2)}{0.001}$

$=52.27814$m/s

(b) $y(1)=56(1)-1.86(1)^2$

$=56-1.86$

$= 54.14$ m/s