+1904 
Anyone who know how to answer this problem and can give step by step instructions, I would really appreciate it. Skip e because the professor said not to do that one.
+128448 I can do a few
a) The velocity is positive on 0 < t < 1
This means the ball is rising during this time
b) The antiderivative is s (t) = 32t - 16t^2 + C
At s(0) we have 32(0) - 16(0)^2 + C = 0
So C = 0
c) s(1) - s(1/2) = [ 32(1) - 16(1)^2 ] - [ 32(1/2) - 16(1/2)^2 ] = [ 16] - [ 16 - 4] = 4
This represents the change in postion from t = 1/2 to t = 1
[ It rises 4 ft in this interval = the displacement on this interval ]
d) At t = (1/2).....the height of the triangle formed is 32 - 32(1/2) = 32 - 16 = 16
So....the area = (1/2) base * height = (1/2) (1/2)(16) = 4
This area also represents the displacement on t =1/2 to t =1
f) s(2) - s(0) = [ 32(2) - 16(2)^2] - 0 = 0
This means that the total displacement = 0 ......it falls as far as it rises
Note that the velocity at t = 0 = 32 ft/s
And at t = 2 = -32 ft/s
This makes sense.....the terminal velocity = the beginning velocity....only in the opposite direction
+118608 A lesson in understanding definite integrations
Think about what you are finding when you get the area under a curve.
In this instance from t=0 to t=1 you are finding the area of a triangle which is 4. But 4 what?
Look at the units. They are t time (seconds) in the horizonal directions and velocity (feet/sec) in the vertical direction.
When you find the area you can find the units the same way. The unit of the area will be \(seconds\times \frac{feet}{sec}=feet\)
So that is why the area under the velocity curve gives displacement.
And youcan find the area with integration.
NOW
Lets think about integration for a moment.
The integration sign. \(\int\) is a stylized S and it stands for SUM
What you are doing with any definite integral is finding the sum of all the rectangles (with infinitely small width) that lie between the curve and the axis that you are integrating with respect to (usually the horizonal axis).
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A lesson in understanding differentiation
With differentiation the d stands for difference. So \(\frac{dx}{dt}=\frac{\text{difference in x}}{\text{difference in t }} \)
Since this is the answer as t tends to 0, dx/dt results in the instantaneous velocity.
The gradient of the tangent to a curve IS the differential to the curve at that point.
If you have questions then please ask them 
