If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, it height in feet after t second is given by y = 70 t - 16 t^2. Find the average velocity for the time period begining when t = 1 and lasting
(i) 0.1 seconds
(ii) 0.01 seconds
(iii) 0.001 seconds
Finally based on the above results, guess what the instantaneous velocity of the ball is when t =1
So, the question is stating that the time period you are looking for is when the ball is thrown at 1 second and goes to .1 second. So, you want the velocity of the ball at the interval [1,1.1]. Velocity is the change in distance/ change in time. What is the change of the distance of the change in time from 1 second to 1.1 second? If the equation of f(x)= 70 t-16 t^2, the f(x) is the distance of the ball, so you should see that f(1.1)-f(1) is the change in height. Everything together is the average rate of change formula, where [f(1.1)-f(1)]/1.1-1=the average velocity of the ball. Now, you can use the formula to make the interval even smaller.
To find the average velocity, we need to use the following formula:
Average velocity = (change in position) / (change in time)
Let's apply this formula to the given scenario:
(i) When the time period is 0.1 seconds:
Initial time (t1) = 1 second
Final time (t2) = 1.1 seconds
Change in time = t2 - t1 = 1.1 - 1 = 0.1 seconds
To find the change in position, we substitute the values of t1 and t2 into the equation y = 70t - 16t^2:
Change in position = y(t2) - y(t1) = (70(1.1) - 16(1.1)^2) - (70(1) - 16(1)^2)
Simplifying the above equation will give us the change in position.
(ii) When the time period is 0.01 seconds:
Initial time (t1) = 1 second
Final time (t2) = 1.01 seconds
Change in time = t2 - t1 = 1.01 - 1 = 0.01 seconds
Using the same equation, y(t2) - y(t1), where y is the height function, we can find the change in position.
(iii) When the time period is 0.001 seconds:
Initial time (t1) = 1 second
Final time (t2) = 1.001 seconds
Change in time = t2 - t1 = 1.001 - 1 = 0.001 seconds
Again, substituting t1 and t2 into the equation y(t2) - y(t1) will give us the change in position.
Based on the average velocities calculated for different time periods, we can estimate the instantaneous velocity of the ball when t = 1 by choosing a smaller time interval. A smaller time interval will yield a more accurate value for the instantaneous velocity.
To find the average velocity for a given time period, we need to calculate the change in height and divide it by the change in time. Let's go through each part step by step:
(i) For a time period of 0.1 seconds:
- We calculate the change in height by plugging in t = 0.1 into the equation and subtracting the height at t = 1:
Δy = (70 * 0.1 - 16 * (0.1)^2) - (70 * 1 - 16 * 1^2)
Δy = (7 - 1.6) - (70 - 16)
Δy = 5.4 - 54
Δy = -48.6 ft
- The change in time is simply 0.1 seconds.
- The average velocity is given by the ratio of the change in height to the change in time:
Average velocity = Δy / change in time
Average velocity = -48.6 ft / 0.1 s
Average velocity = -486 ft/s
(ii) For a time period of 0.01 seconds, follow the same process:
Δy = (70 * 0.01 - 16 * (0.01)^2) - (70 * 1 - 16 * 1^2)
Δy = (0.7 - 0.0016) - (70 - 16)
Δy = 0.6984 - 54
Δy = -53.3016 ft
Average velocity = -53.3016 ft / 0.01 s
Average velocity = -5330.16 ft/s
(iii) For a time period of 0.001 seconds:
Δy = (70 * 0.001 - 16 * (0.001)^2) - (70 * 1 - 16 * 1^2)
Δy = (0.07 - 0.000016) - (70 - 16)
Δy = 0.069984 - 54
Δy = -53.930016 ft
Average velocity = -53.930016 ft / 0.001 s
Average velocity = -53,930.016 ft/s
Based on the above results, we can see that the average velocity is decreasing as the time period becomes shorter.
To estimate the instantaneous velocity at t = 1, we can take the limit as the time period approaches zero. As the time period becomes infinitesimally small, the average velocity converges to the instantaneous velocity at that specific moment. Therefore, we can estimate that the instantaneous velocity of the ball when t = 1 is approximately -54,000 ft/s.