calc1 - average velocity?

Question

Not too sure on calculating average velocity, here is a question

if a ball is thrown into the air with a velocity of 40ft/s its height in feet t seconds later is given by y=40t-16t^2

find the average velocity for the time period beginning when t=2 and lasting

a) .5 sec
b) .1 sec

please show work

Answer 1

Well, this isnt too difficult... its just a lot of work.

The average velocity is what? Its the velocity at the beginning... plus the velocity at the end... divided by two.

Pretty easy there.

You already know the time it took to make the journey from beginning to end... its given to you (0.5 sec and 0.1 sec)... and it gives you initial time at t=2 seconds.

The problem now is finding the velocity at the beginning and at the end of each trip.

You have a function defining height per unit of time. What is velocity besides a change in height over a change in time? Hmmm... This is a differentiation problem.

So, derive your height over time function to create a velocity over time function.

y' = d/dt 40t - 16 t^2 = 40 - 32 t

So, at any given time, t, the velocity is given by the function y'.

Amazing!

Finding the velocity at t=2...
y' = 40 - 32 (2) = -24 ft/sec.

Remember, velocity is a vector based measurement. A negative value means that it is falling towards the earth... moving in the opposite direction it was initially launched at.

a) 0.5 seconds later, t will equal 2 seconds plus 0.5 seconds... for at total of 2.5 seconds into the balls launch.

b) 0.1 seconds later, t will equal 2 seconds plus 0.1 seconds... for at total of 2.1 seconds into the balls launch.

So, the velocities at times t=2.5 and t=2.1 seconds are found just as easily.

y' = 40 - 32 (2.1) = -27.2 ft/sec.
y' = 40 - 32 (2.5) = -40 ft/sec.

Now, you know both the start velocities and end velocities for the ball during the various time periods.

The average is merely the start velocity (at t=2.0 sec) added to each of the other two velocities (at t=2.1 and at t=2.5)... divided by two.

From 2.0 to 2.1 seconds, we go from -24 ft/sec to -27.2 ft/sec. That is an average of (-24 ft/sec + -27.2 ft/sec)/2 = -25.6 ft/sec

From 2.0 to 2.5 seconds, we go from -24 ft/sec to -40 ft/sec. That is an average of (-24 ft/sec + -40 ft/sec)/2 = -32 ft/sec

Answer 2

Average speed is defined as half of the final speed. Velocity implies the direction . So to define average velocity with a moving object, let say an airplain for example whcih changes direction continuously,you have to add all the directions vectorialy which would give an average direction (the resultant sum of all directions) So the velocity average would be the average speed at an average direction. This yields a vector of which speed is the magnetude and the angle is the direction.

Answer 3

Average velocity is the distance traveled divided by the time elapsed. Therefore, the average velocity from time 2 seconds to 2.5 seconds is (y(2.5) - y(2))/.5 or
{40(2.5) - 16((2.5)^2) - [40(2) - 16(2^2)]}/.5 = -32. The negative sign indicates that the ball is on it way down over this time interval.

Similar for part b.