Asked by Nick
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.)
What is 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]
y = 52t − 1.86t2.
(Round your answers to two decimal places.)
What is 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]
Answers
Answered by
Steve
the average speed is the total distance divided by the time. The total distance is the integral of the velocity. So, for
(i) the total distance traveled is
∫[1,2] (52t − 1.86t^2) dt = 73.66 meters
since the interval length is 1 second, that is the average speed.
do the others in like wise.
(i) the total distance traveled is
∫[1,2] (52t − 1.86t^2) dt = 73.66 meters
since the interval length is 1 second, that is the average speed.
do the others in like wise.
Answered by
Steve
Oops. My bad. I thought the equation was velocity. Since it is distance, the average speed is just the distance covered, divided by the time (y(b)-y(a))/(b-a)
For (i), then we have
y(2) = 96.56
y(1) = 50.14
so, the average speed is (96.56-50.14)/(2-1) = 46.42 m/s
For (i), then we have
y(2) = 96.56
y(1) = 50.14
so, the average speed is (96.56-50.14)/(2-1) = 46.42 m/s
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