1.if a ball is thrown into the air with a velocity of 35ft/s, it's height in feet after t seconds is given by y=35t-10t^2. Find the averag velocity for the time period beginning when t=4 and lasting 0.05s.
2.if a ball is thrown into the air with a velocity of 30ft/s, it's height in feet after t seconds is given by y=30t-20t^2. Find the instantaneous velocity when t=3
Last time I was on this earth, gravity would cause an equation of the type
y = -16t^2 .....
So assuming you have a typo and it should have been
y = 35t - 16t^2
Second problem:
the ball would not be in the air for 4 seconds,
so your whole first question is actually just gibberish.
Once you fix it, find y when t = 4
find y when t = 4.05
evaluate: ( y(4.05) - y(4) )/.05
2. same thing here,
after 3 seconds the ball has reached the ground, and no longer has any velocity
if we ignore reality,
dy/dt = 30 - 40t
when t = 3
dy/dt = 30 - 120 = -90 ft/s
1.
when t=4, y = 35(4) - 16(16) = -20
There is no error in the problem.
I see the typo as specifying feet. In metric units, g=9.8, or roughly 10, so Jacke, redo the calculations, with g=10 and get the correct answer.
Is it ok if I get a negative answer???
OK thanks I got it
To find the average velocity for the time period in question, you need to calculate the change in position (height) divided by the change in time.
1. Average Velocity:
The formula to calculate the average velocity is (change in position) / (change in time).
In this case, the position is given by the equation y = 35t - 10t^2, and the time period of interest is from t = 4 to t = 4 + 0.05 = 4.05 seconds.
To find the change in position, substitute the starting time (t = 4) into the equation:
y1 = 35(4) - 10(4)^2 = 140 - 160 = -20 feet.
Next, substitute the ending time (t = 4.05) into the equation:
y2 = 35(4.05) - 10(4.05)^2 = 141.75 - 164.025 = -22.275 feet.
Now, calculate the change in position:
Change in position = y2 - y1 = (-22.275) - (-20) = -2.275 feet.
The change in time is given by subtracting the starting time from the ending time:
Change in time = 4.05 - 4 = 0.05 seconds.
Finally, divide the change in position by the change in time to find the average velocity:
Average velocity = (change in position) / (change in time) = -2.275 / 0.05 = -45.5 ft/s.
Therefore, the average velocity for the time period starting at t = 4 and lasting for 0.05 seconds is -45.5 ft/s.
2. Instantaneous Velocity:
To find the instantaneous velocity at a specific time (t = 3), you need to calculate the derivative of the position equation with respect to time.
The position equation is y = 30t - 20t^2. Take the derivative of this equation with respect to time:
dy/dt = 30 - 40t.
To find the instantaneous velocity, substitute t = 3 into the derivative equation:
dy/dt = 30 - 40(3) = 30 - 120 = -90 ft/s.
Therefore, the instantaneous velocity when t = 3 is -90 ft/s.