The distance (in feet) of an object from a point is given by s(t)=3t2−5, where t is in seconds. Round your answer to two decimal places.
1. What is the average velocity of the object between t=1 and t=8.5?
2. By using smaller and smaller intervals around 1, estimate the instantaneous velocity at time t=1.
(tried several different ways and can't seem to get it right! Appreciate the help in advance!!)
s(t) = 3t^2 − 5
I assume you mean location is s, not distance, because it is negative at t = 0
s(8.5) = 3* 8.5^2 - 5 = 211.75
s(1) = 3-5 = -2
distance moved in 7.5 seconds = 213.75
average velocity = 213.75 / 7.5 = 28.2 ft/s
for part b I will take the derivative
ds/dt = 6 t
at t = 1, velocity = 6 * 1 = 6 ft/s
your way"
t = 1
s = 3 * 1^2 - 5 = -2
at t = 1.1
s = 3 *1.1^2 - 5 = -1.37
av speed = (-1.37 - -2) / 0.1
= 0.63 /9.1 = 6.3
well the real answer is 6 so close
now try t = 1.01 :)
for all such average problems, use the difference quotient
(f(t+h) - f(t)) / h
If you think you're stuck, show some work.
To find the average velocity between t=1 and t=8.5, we can use the formula:
Average Velocity = (Change in Distance) / (Change in Time)
1. Average Velocity between t=1 and t=8.5:
s(8.5) = 3(8.5)^2 - 5 = 232.25 feet
s(1) = 3(1)^2 - 5 = -2 feet
Change in Distance = s(8.5) - s(1) = 232.25 - (-2) = 234.25 feet
Change in Time = 8.5 - 1 = 7.5 seconds
Average Velocity = 234.25 / 7.5 ≈ 31.23 feet/second
2. To estimate the instantaneous velocity at t=1, we can use smaller intervals around t=1. Let's choose an interval of 0.1 seconds before and after t=1.
s(1.1) = 3(1.1)^2 - 5 = -1.69 feet
s(0.9) = 3(0.9)^2 - 5 = -2.79 feet
Change in Distance = s(1.1) - s(0.9) = -1.69 - (-2.79) = 1.1 feet
Change in Time = 1.1 - 0.9 = 0.2 seconds
Instantaneous Velocity ≈ Change in Distance / Change in Time ≈ 1.1 / 0.2 ≈ 5.5 feet/second
Therefore, the estimate of the instantaneous velocity at t=1 is approximately 5.5 feet/second.
To find the average velocity of the object between t = 1 and t = 8.5, you need to calculate the change in distance divided by the change in time.
1. Average velocity between t = 1 and t = 8.5:
To find the change in distance, subtract the distance at t = 1 from the distance at t = 8.5:
s(8.5) - s(1) = (3(8.5)^2 - 5) - (3(1)^2 - 5)
Evaluate this expression:
s(8.5) - s(1) = (3(72.25) - 5) - (3 - 5)
s(8.5) - s(1) = (216.75 - 5) - (-2)
s(8.5) - s(1) = (211.75) - (-2)
s(8.5) - s(1) = 213.75
Now, find the change in time by subtracting the initial time from the final time:
8.5 - 1 = 7.5
Finally, calculate the average velocity by dividing the change in distance by the change in time:
average velocity = (distance change) / (time change)
average velocity = 213.75 / 7.5
average velocity ≈ 28.5 ft/s (rounded to two decimal places)
2. Estimating instantaneous velocity at t = 1:
To estimate the instantaneous velocity at t = 1, you need to take smaller and smaller intervals around t = 1. We can use a numerical approach to estimate it.
Choose smaller intervals around t = 1, such as t = 1.1, t = 1.01, t = 1.001, and so on. The smaller the interval, the closer the instantaneous velocity estimate will be to the actual value.
For example, let's calculate the average velocity between t = 1 and t = 1.1:
s(1.1) - s(1) = (3(1.1)^2 - 5) - (3(1)^2 - 5)
Evaluate this expression:
s(1.1) - s(1) = (3(1.21) - 5) - (3 - 5)
s(1.1) - s(1) = (3.63 - 5) - (-2)
s(1.1) - s(1) = (3.63) - (-2)
s(1.1) - s(1) = 5.63
Now, find the change in time:
1.1 - 1 = 0.1
Calculate the average velocity:
average velocity = (distance change) / (time change)
average velocity = 5.63 / 0.1
average velocity = 56.3 ft/s (rounded to two decimal places)
Repeat this process with smaller intervals if you want a more accurate estimate of the instantaneous velocity at t = 1.