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]

Just posted a question like this but put in 54 instead of 52. Please show approbate steps. Idk where to even start

average speed is ... distance / time

find the height at the beginning and end of the interval , then divide the height change by the time change

To find the average speed over a given time interval, we need to calculate the total distance traveled during that interval and divide it by the duration of the interval.

We can start by finding the distance traveled between two moments in time t1 and t2 using the given equation: y = 52t - 1.86t^2.

(i) [1, 2]:
First, let's find the distance traveled between t = 1 and t = 2.
For t = 1, we have y = 52(1) - 1.86(1)^2 = 50.14 meters.
For t = 2, we have y = 52(2) - 1.86(2)^2 = 52 - 1.86(4) = 52 - 7.44 = 44.56 meters.

Therefore, the distance traveled between t = 1 and t = 2 is 44.56 - 50.14 = -5.58 meters (negative sign indicating upward motion).

The average speed is then -5.58 meters divided by the time interval of 2 - 1 = 1 second, which equals -5.58 m/s.

(ii) [1, 1.5]:
To find the distance traveled between t = 1 and t = 1.5:
For t = 1, we have y = 52(1) - 1.86(1)^2 = 50.14 meters.
For t = 1.5, we have y = 52(1.5) - 1.86(1.5)^2 = 52(1.5) - 2.79 = 78 - 2.79 = 75.21 meters.

The distance traveled between t = 1 and t = 1.5 is given by 75.21 - 50.14 = 25.07 meters.

The average speed would be 25.07 meters divided by the time interval of 1.5 - 1 = 0.5 seconds, which equals 50.14 m/s.

(iii) [1, 1.1]:
The distance traveled between t = 1 and t = 1.1 is found as follows:
For t = 1, we have y = 50.14 meters.
For t = 1.1, we have y = 52(1.1) - 1.86(1.1)^2 = 57.2 - 2.0366 = 55.1634 meters.

The distance traveled between t = 1 and t = 1.1 is 55.1634 - 50.14 = 5.0234 meters.

The average speed is 5.0234 meters divided by the time interval of 1.1 - 1 = 0.1 seconds, resulting in 50.23 m/s.

(iv) [1, 1.01]:
The distance traveled between t = 1 and t = 1.01 is calculated as follows:
For t = 1, we have y = 50.14 meters.
For t = 1.01, we have y = 52(1.01) - 1.86(1.01)^2 = 52.52 - 2.354346 = 50.165654 meters.

The distance traveled between t = 1 and t = 1.01 is 50.165654 - 50.14 = 0.025654 meters.

The average speed is the distance of 0.025654 meters divided by the time interval of 1.01 - 1 = 0.01 seconds, giving us 2.5654 m/s.

(v) [1, 1.001]:
Lastly, let's find the distance traveled between t = 1 and t = 1.001:
For t = 1, we have y = 50.14 meters.
For t = 1.001, we have y = 52(1.001) - 1.86(1.001)^2 = 52.052 - 2.068214 = 49.983786 meters.

The distance traveled between t = 1 and t = 1.001 is 49.983786 - 50.14 = -0.156214 meters (indicating upward motion).

The average speed is the distance of -0.156214 meters divided by the time interval of 1.001 - 1 = 0.001 seconds, which gives us -156.214 m/s.

In summary, the average speeds over the given time intervals are:
(i) [1, 2]: -5.58 m/s
(ii) [1, 1.5]: 50.14 m/s
(iii) [1, 1.1]: 50.23 m/s
(iv) [1, 1.01]: 2.57 m/s
(v) [1, 1.001]: -156.21 m/s