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]
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
To determine the average speed over each time interval, we need to find the total distance traveled within that interval and divide it by the duration of the interval.
(i) [1, 2]:
To find the average speed over the interval [1, 2], we need to calculate the total distance traveled and divide it by the duration of the interval.
First, let's find the height at t = 1 and t = 2 using the equation y = 52t - 1.86t^2.
At t = 1:
y = 52(1) - 1.86(1)^2 = 52 - 1.86 = 50.14 m
At t = 2:
y = 52(2) - 1.86(2)^2 = 104 - 7.44 = 96.56 m
The total distance traveled between t = 1 and t = 2 is the difference in heights: 96.56 - 50.14 = 46.42 m.
The duration of the interval is 2 - 1 = 1 second.
Average speed = Total distance traveled / Duration of interval = 46.42 m / 1 s = 46.42 m/s.
(ii) [1, 1.5]:
Following the same process, we find:
At t = 1:
y = 52(1) - 1.86(1)^2 = 50.14 m
At t = 1.5:
y = 52(1.5) - 1.86(1.5)^2 = 78 - 5.29 = 72.71 m
The total distance traveled between t = 1 and t = 1.5 is the difference in heights: 72.71 - 50.14 = 22.57 m.
The duration of the interval is 1.5 - 1 = 0.5 seconds.
Average speed = 22.57 m / 0.5 s = 45.14 m/s.
(iii) [1, 1.1]:
Using the same approach:
At t = 1:
y = 52(1) - 1.86(1)^2 = 50.14 m
At t = 1.1:
y = 52(1.1) - 1.86(1.1)^2 = 57.2 - 2.36 = 54.84 m
The total distance traveled between t = 1 and t = 1.1 is the difference in heights: 54.84 - 50.14 = 4.70 m.
The duration of the interval is 1.1 - 1 = 0.1 seconds.
Average speed = 4.70 m / 0.1 s = 47.00 m/s.
(iv) [1, 1.01]:
Similarly:
At t = 1:
y = 52(1) - 1.86(1)^2 = 50.14 m
At t = 1.01:
y = 52(1.01) - 1.86(1.01)^2 = 50.14 + 0.526 - 0.0374 = 50.63 m
The total distance traveled between t = 1 and t = 1.01 is the difference in heights: 50.63 - 50.14 = 0.49 m.
The duration of the interval is 1.01 - 1 = 0.01 seconds.
Average speed = 0.49 m / 0.01 s = 49.00 m/s.
(v) [1, 1.001]:
Applying the same process:
At t = 1:
y = 52(1) - 1.86(1)^2 = 50.14 m
At t = 1.001:
y = 52(1.001) - 1.86(1.001)^2 = 50.14 + 0.052 - 0.00371 = 50.19 m
The total distance traveled between t = 1 and t = 1.001 is the difference in heights: 50.19 - 50.14 = 0.05 m.
The duration of the interval is 1.001 - 1 = 0.001 seconds.
Average speed = 0.05 m / 0.001 s = 50.00 m/s.
In summary:
(i) [1, 2]: Average speed = 46.42 m/s
(ii) [1, 1.5]: Average speed = 45.14 m/s
(iii) [1, 1.1]: Average speed = 47.00 m/s
(iv) [1, 1.01]: Average speed = 49.00 m/s
(v) [1, 1.001]: Average speed = 50.00 m/s.
To find the average speed over a given time interval, we need to calculate the change in position over that time interval, and then divide it by the duration of the interval.
The change in position can be obtained by finding the difference between the initial height and the final height at the end of the time interval.
Let's calculate the average speed for each given time interval:
(i) [1, 2]:
To find the initial and final heights at t = 1 and t = 2, substitute these values into the equation:
Initial height (t=1) = y(1) = 52(1) - 1.86(1^2) = 50.14
Final height (t=2) = y(2) = 52(2) - 1.86(2^2) = 46.48
Change in position = Final height - Initial height = 46.48 - 50.14 = -3.66
Average speed = Change in position / Time interval = -3.66 / (2 - 1) = -3.66 m/s
(ii) [1, 1.5]:
Initial height (t=1) = y(1) = 52(1) - 1.86(1^2) = 50.14
Final height (t=1.5) = y(1.5) = 52(1.5) - 1.86(1.5^2) = 50.27
Change in position = Final height - Initial height = 50.27 - 50.14 = 0.13
Average speed = Change in position / Time interval = 0.13 / (1.5 - 1) = 0.13 m/s
(iii) [1, 1.1]:
Initial height (t=1) = y(1) = 52(1) - 1.86(1^2) = 50.14
Final height (t=1.1) = y(1.1) = 52(1.1) - 1.86(1.1^2) = 50.06
Change in position = Final height - Initial height = 50.06 - 50.14 = -0.08
Average speed = Change in position / Time interval = -0.08 / (1.1 - 1) = -0.08 m/s
(iv) [1, 1.01]:
Initial height (t=1) = y(1) = 52(1) - 1.86(1^2) = 50.14
Final height (t=1.01) = y(1.01) = 52(1.01) - 1.86(1.01^2) = 50.13
Change in position = Final height - Initial height = 50.13 - 50.14 = -0.01
Average speed = Change in position / Time interval = -0.01 / (1.01 - 1) = -0.01 m/s
(v) [1, 1.001]:
Initial height (t=1) = y(1) = 52(1) - 1.86(1^2) = 50.14
Final height (t=1.001) = y(1.001) = 52(1.001) - 1.86(1.001^2) = 50.14
Change in position = Final height - Initial height = 50.14 - 50.14 = 0
Average speed = Change in position / Time interval = 0 / (1.001 - 1) = 0 m/s
Therefore, the average speeds for the given time intervals are:
(i) [1, 2] = -3.66 m/s
(ii) [1, 1.5] = 0.13 m/s
(iii) [1, 1.1] = -0.08 m/s
(iv) [1, 1.01] = -0.01 m/s
(v) [1, 1.001] = 0 m/s
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.