Jaden Harris' Velocity Ques. at Yahoo Answers

[MarkFL]

Here is the question:

Average Velocity and instantaneous velocity?

I am having so much trouble with this question, and I have no idea why. I know that average velocity is based upon a secant and is basically a slope formula. Instantaneous velocity is based upon a tangent and is basically a point. However, this question is confusing me, maybe I'm staying up too late. Help?

When a ball is thrown vertically upward into the air with a velocity of 72 ft/sec its height, y(t), in feet after t seconds is given by y(t) = 72t - 16t^2. Find the average velocity of the ball over the interval from 4 to 4+h seconds, h does not = 0.
a) Avg. Vel.= -(57-16h) ft/sec
b)Avg. Vel.= -(57+h) ft/sec
c) Avg. Vel.= -(57+16h) ft/sec
d) Avg. Vel.= -(57-h) ft/sec
e) Avg. Vel.= -(56+16h) ft/sec
f) Avg. Vel.= -(56-16h) ft/sec

Then it proceeds to ask:
Find the instantaneous velocity of the ball after 4 seconds:
a) Instantaneous Vel. =-56 ft/sec
b) Instantaneous Vel. =-55 ft/sec
c) Instantaneous Vel. =-53 ft/sec
d) Instantaneous Vel. =-54 ft/sec
e) Instantaneous Vel. =-57 ft/sec

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Jaden Harris,

For the first part of the question, we need to compute the difference quotient:

\(\displaystyle \frac{\Delta y}{\Delta t}=\frac{y(4+h)-y(h)}{h}\,\frac{\text{ft}}{\text{s}}\)

So, we find:

\(\displaystyle y(4+h)-y(h)=\left(72(4+h)-16(4+h)^2 \right)-\left(72(4)-16(4)^2 \right)=\)

\(\displaystyle 72(4)+72(h)-16\left(16+8h+h^2 \right)-72(4)+16(16)=72h-128h-16h^2=-8h(7+2h)\)

Hence:

\(\displaystyle \frac{\Delta y}{\Delta t}=\frac{-8h(7+2h)}{h}\,\frac{\text{ft}}{\text{s}}=-8(7+2h)\,\frac{\text{ft}}{\text{s}}=-(56+16h)\,\frac{\text{ft}}{\text{s}}\)

Thus, e) is the correct answer.

For the second part of the question, we wish to evaluate the limit:

\(\displaystyle y'(4)\equiv\lim_{h\to0}-(56+16h)\,\frac{\text{ft}}{\text{s}}=-56\,\frac{\text{ft}}{\text{s}}\)

Thus, a) is the correct answer.