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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.

User Tim Tran
by
5.4k points

1 Answer

2 votes

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

Explanation:

The average speed over the given time interval [a,b]: =
(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 =
(y(2)-y(1))/(2-1)


=(56(2)-1.86(2)^2-(56(1)-1.86(1)^2))/(1)


=50.42 m/s

(ii) average speed =
(y(1.5)-y(1))/(1.5-1)


=(56(1.5)-1.86(1.5)^2-(56(1)-1.86(1)^2))/(0.5)


=51.35m/s

(iii) average speed =
(y(1.1)-y(1))/(1.1-1)


=(56(1.1)-1.86(1.1)^2-(56(1)-1.86(1)^2))/(0.1)


=52.094m/s

(iv) average speed =
(y(1.01)-y(1))/(1.01-01)


=(56(1.01)-1.86(1.01)^2-(56(1)-1.86(1)^2))/(0.01)


=52.2614m/s

(v) average speed =
(y(1.001)-y(1))/(1.001-01)


=(56(1.001)-1.86(1.001)^2-(56(1)-1.86(1)^2))/(0.001)


=52.27814m/s

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


=56-1.86


= 54.14 m/s

User Brigadeiro
by
4.7k points
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