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.)
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
what is the question?
Find the average speed over the given time intervals.
ok, avg speed=
=(y(finaltime)-y(initialtime)) divided by(finaltime-initialtime)
y = 52t − 1.86t2.
lets do iv)
y(1)=52-1.86=51.14
y(1.001)=52.052-1.86(1.001)^2=50.188
so avg apeed= (.188)/.001=188m/s
To find the height of the arrow at a specific time, we need to substitute the given time value into the equation y = 52t - 1.86t^2.
(i) For the time interval [1, 2]:
- Substitute t = 1 into the equation: y = 52(1) - 1.86(1)^2 = 50.14 meters. Round to two decimal places.
- Substitute t = 2 into the equation: y = 52(2) - 1.86(2)^2 = 102.24 meters. Round to two decimal places.
Therefore, the height of the arrow in meters during the time interval [1, 2] is between 50.14 and 102.24 meters.
(ii) For the time interval [1, 1.5]:
- Substitute t = 1 into the equation: y = 52(1) - 1.86(1)^2 = 50.14 meters. Round to two decimal places.
- Substitute t = 1.5 into the equation: y = 52(1.5) - 1.86(1.5)^2 = 70.92 meters. Round to two decimal places.
Therefore, the height of the arrow in meters during the time interval [1, 1.5] is between 50.14 and 70.92 meters.
(iii) For the time interval [1, 1.1]:
- Substitute t = 1 into the equation: y = 52(1) - 1.86(1)^2 = 50.14 meters. Round to two decimal places.
- Substitute t = 1.1 into the equation: y = 52(1.1) - 1.86(1.1)^2 = 54.61 meters. Round to two decimal places.
Therefore, the height of the arrow in meters during the time interval [1, 1.1] is between 50.14 and 54.61 meters.
(iv) For the time interval [1, 1.01]:
- Substitute t = 1 into the equation: y = 52(1) - 1.86(1)^2 = 50.14 meters. Round to two decimal places.
- Substitute t = 1.01 into the equation: y = 52(1.01) - 1.86(1.01)^2 = 50.58 meters. Round to two decimal places.
Therefore, the height of the arrow in meters during the time interval [1, 1.01] is between 50.14 and 50.58 meters.
(v) For the time interval [1, 1.001]:
- Substitute t = 1 into the equation: y = 52(1) - 1.86(1)^2 = 50.14 meters. Round to two decimal places.
- Substitute t = 1.001 into the equation: y = 52(1.001) - 1.86(1.001)^2 = 50.16 meters. Round to two decimal places.
Therefore, the height of the arrow in meters during the time interval [1, 1.001] is between 50.14 and 50.16 meters.
Overall, we can conclude that the height of the arrow increases as time passes, but it reaches a maximum point before starting to descend.