calculating average velocity
When a ball is thrown vertically upward into the air with a velocity of 56 ft/sec its height, y(t), in feet after t seconds is given by \[y(t)=56t-16t ^{2}\] Find the average velocity of the ball over the interval from 3 to 3+h seconds, \[h \neq0\]
answer choices: 1. average vel. = −(41 + 16h) ft/sec 2. average vel. = −(40 − h) ft/sec 3. average vel. = −(41 − 16h) ft/sec 4. average vel. = −(41 + h) ft/sec 5. average vel. = −(40 + 16h) ft/sec 6. average vel. = −(40 − 16h) ft/sec
average velocity = \(\large \frac{y(3+h) - y(3)}{(3+h) - h}\)
simplify
@ganeshie8 I got -(40+16h) ft/sec
Now..how do I find the instantaneous velocity of the ball after 3 seconds?
Excellent ! thats correct
familiar wid calculus ?
haha I am taking calculus.
would I substitute t for 3 from the answer I got from the first problem?
wait..no that's not rigt.
no, that just gives u height
oh i see. what do I do, now?
did u get how the average velocity was found ?
Average velocity = \(\large \frac{y(3+h) - y(3)}{(3+h) - h}\) u can use the same formula for instantaneous velocity, but u need to MAKE THE h SOOO SMALL
then oly, u wil get the velocoty right after 3 seconds. getting ?
yeah I understood that formula
that's what I used earlier
make h=0.001 and see wat u wud get
Do i need to plug in a 3 anywhere?
3 is already there in the formula, just put h value and simplify
I seee. thank you!
wait a sec, you have already simplifies the average velocity average velocity between 3 & h seconds = -(40+16h) ft/sec
right ?, now simply put h = 0.001 above. you will get instantaneous velocity at time 3
that wud be oly the APPROXIMATE value
wait. what if I just plug in 3 into \[56t-16t ^{2}\]
@ganeshie8
@ganeshie8 is the answer -40?
yes thats right
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