Question

Please Check, getting weird answer.
object is thrown in the air [...] modelled by h(t)=-5t^2+20t+1 [...] Find Instantaneous Velocity at t= 2
h(t+h) - h(t) / h

Answer 1

h(t) = 21 [plugged in t=2]
h(2+h) = -5(2+h)^2 +20(2+h) + 1
(25h^2 +100h+100) + 40 +20h + 1
h(2+h)=25h^2+120h+141
h(2+h)-h(2) / h
25h^2+120h+141 - 21 / h
h(25h + 120 +140) -21 /h
25h+240
25(0) + 240
=240m/s

Answer 2

should be 141^

Answer 3

Not the answer in the line " h(25h +120 +141**) - 21 / h

Answer 4

the instantaneous speed is the limit of the expression as h approaches zero

Answer 5

where h is a small change in t

Answer 6

when you expanded " -5(2+h)^2" from the 2nd line to the 3rd line, you made an error...

Answer 7

Rather than subbing in a small value for h. you can sub in 0 = h because of simplification. Therefore you can find the slope of the secant between (2 + 0) and h(2+0) Which really is the slope at a single point

Answer 8

Jake I noticed I made an error, I have a second post stating that it should be 141 not 140

Answer 9

shouldn't -5(2+h)^2 simplify to 5(4 + 4h + h^2) = 20 + 20h + 5h^2

Answer 10

I thought it would be (-10 -5h)(-10-5h) then expand.

Answer 11

that PEMDA order of operations thing says you have to do exponents before you can do the multiplication.

Answer 12

I see what you were thinking, but you need to square it first, then distribute the 5 across