Question

f an arrow is shot upward on the moon with a speed of 52 m/s, its height in meters t seconds later is given by
y = 52t ? 0.83t2.
(Round your answers to two decimal places.)
(a) Find the average speed over the given time intervals.
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
(b) Estimate the speed when t = 1.

How does one go about doing this?

Answer 1

Sorry!!!
a)v = dy/dt = 52-(2*0.83*t)
So average velocity:
(i) [1, 2] - 1.66
(ii) [1, 1.5] - 1.66
(iii) [1, 1.1] - 1.66
(iv) [1, 1.01] - 1.66
(v) [1, 1.001] - 1.66
b)the speed when t=1 is 50.34m/sec

Answer 2

a)v = dy/dt = 52-(2*0.83*t)
So average velocity:
(i) [1, 2] - 0.83
(ii) [1, 1.5] - 0.83
(iii) [1, 1.1] - 0.83
(iv) [1, 1.01] - 0.83
(v) [1, 1.001] - 0.83
b)the speed when t=1 is 50.34m/sec

Answer 3

y = 52t - 0.83t^2
dy/dt = speed, = -0.83*2*t = 1.66t
(I) [1,2] t=1 speed =1.66 m/s , t=2 speed = 3.32 average = 2.49 m/s
(II)[1,1.5] t =1 speed = 1.66m/s, t=1.5 speed =2.49 m/s average = 2.07 m/s
(III)[1,1.1] t=1 speed=1.66 m/s, t=1.1 speed = 1.826 m/s average = 1.74 m/s
(V)[1,1.001] t=1 speed = 1.66 m/s, t=1.001 speed = 1.66166 m/s average = 1.66 m/s
(V)[1,1.01] t=1 speed =1.66 m/s. t=1.01 speed = 1.6766 m/s average = 1.67 m/s

b) at t=1 speed = 1.66 m/s

Answer 4

a)v = dy/dt = 52-(2*0.83*t)
So average velocity:
(i) [1, 2] - 1.66
(ii) [1, 1.5] - 0.83
(iii) [1, 1.1] - 0.166
(iv) [1, 1.01] - 0.0166
(v) [1, 1.001] - 0.00166
b)the speed when t=1 is 50.34m/sec