Question

can someone actually explain this problem for me?

Answer 1

A ball is dropped from a tower 350 meters above the ground with position function s(t) = 4.9t^2+ 350.

What is the velocity of the ball after 2 seconds?

Include units in your answer.

Answer 2

Velocity is change in displacement with respect to time, right? Therefore you need to take derivative of s(t) in respect to t, to get equation for velocity.

Answer 3

Then you just plug in 2, as t=2s then solve it.

Answer 4

@ganeshie8

Answer 5

@waterineyes

Answer 6

Why to worry when \(\phi\) is here.. :)

Answer 7

How about this:
the change in distance at t=2 and at t+h (h is a tiny number)
is s(t+h) - s(2) (over the time interval (t+h) - t = h , we moved that far
the speed is change in distance divided by change in time:
\[ \frac{s(t+h) - s(2) }{h} \]
if we let h approach 0 (get *very* small) we can get a very accurate speed at *very* near t=2:
\[ \lim_{h\rightarrow 0} \frac{s(t+h) - s(2) }{h}\]

following ?

Answer 8

kinda

Answer 9

I should write that
\[ \lim_{h\rightarrow 0} \frac{s(2+h) - s(2) }{h} \]

Answer 10

does it makes sense that we can find where we are at t=2 using
s(2) ?
and where we are "h" seconds later is s(2+h) ?

Answer 11

yeah

Answer 12

position function s(t) means we put in a time (for example t=2) and we get out a position
(a number telling us how high off the ground we are in meters)
example, when t=0, we would get
s(0)= -4.9*0*0+ 350
s(0)= 350 meters (above the ground)

at t=2 we would be
s(2) = -4.9*2*2 +350 = 330.4 meters above the ground

btw, I assume the equation has a minus sign, because I think we are falling down, and the height should be getting less.

Answer 13

but there is no minus sign..

Answer 14

I would go on the assumption it should be
s(t) = -4.9 t^2 + 350

and work the problem. Then ask your teacher about it.

meanwhile, does this part make sense:
at t=2 we would be s(2) = -4.9*2*2 +350 = 330.4 meters above the ground

Answer 15

i just asked, he said my equation is correct

Answer 16

well, it does not matter except balls usually fall down.
This one will move up (?!)

Using
s(t) = +4.9 t^2 + 350
at t=2 the position (above the ground) will be
s(2) = 4.9*2*2 + 350 = 369.6

ok on that ?

Answer 17

yes!

Answer 18

to find the speed, we need distance traveled divided by amount of time.
Speed = distance/time

Answer 19

Let me restate that
to find the new position after h seconds we find
s(2+h)

then to find the distance we moved, we do
s(2+h) - s(2)

but we need to find s(2+h)
can you do that ?

Answer 20

s(4)

Answer 21

h is 2 right?

Answer 22

s(4) is a position at a later time (with h=2)

but we will want to use algebra and let h be anything. (It means a *tiny* amount)

so we have to put (2+h) into the formula for s in place of t, and then expand using algebra.

can you try to do that ?

Answer 23

what is s

Answer 24

The first step is write down the formula
s(t) = 4.9 t^2 + 350

s(t) is a formula that tells us our position (when we put in a number for t , the time)

Answer 25

when t is 2+h (where h is a tiny number which we will not specify yet)
we would replace t with (2+h) in the formula
s(t) = 4.9 t^2 + 350

to replace the "t", erase it, and write in its place (2+h)
can you do that ?

Answer 26

4.9(2+h)^2+350

Answer 27

yes
now use FOIL to exand (2+h)^2 = (2+h)*(2+h)
can you do that ?

Answer 28

*expand i.e. multiply out

Answer 29

4+2h+2h+h^2

Answer 30

and 2h + 2h can be simplified to 4h
so we now have
s(2+h) = 4.9(2+h)^2+350
s(2+h) = 4.9(4+4h+h^2) + 350

now distribute the 4.9
what do we get ?

Answer 31

19.6+19.6h+19.6h^2+350

Answer 32

ok , so here is what we have:
at t=2 the position s(2) = 350 + 19.6 = 369.6
at t= 2+h the position is at s(2+h) = 19.6+19.6h+19.6h^2+350

ok so far?

Answer 33

yes

Answer 34

any idea how to figure out how far we moved between t=2 and t=2+h ?

Answer 35

uhh..

Answer 36

think about a simpler problem to see if you can get the idea
let's say at t=2 you were at 10 and at t+h you were at 15
how far did you move? How did you figure it out?

Answer 37

5, subtracted

Answer 38

yes. you do position at 2+h minus position at 2

do the same for our positions. can you try that ?

Answer 39

ohhh obviously, such an easy question

Answer 40

youre just saying that i minus h? i dont get what youre asking

Answer 41

s(2+h) is where we are at t=2+h
it is the messy expression s(2+h) = 19.6+19.6h+19.6h^2+350
s(2) is 350+19.6

to find the distance we moved we find the difference s(2+h) - s(2)

(this is how you solved 10 and 15, you did 15-10

Answer 42

is this problem supposed to be this confusing?,...

Answer 43

the idea is that we have two positions and we want to find the difference between them.

This is one of those problems that takes time to work through.

Answer 44

okk

Answer 45

Don't worry about how ugly the expressions seem to be. Can you subtract them ?

Answer 46

First step, write it down

Answer 47

19.6h+19.6h^2

Answer 48

yes

Answer 49

though it looks a bit weird, that is how far we moved between t=2 and t=2+h

how much time is there between t=2 and 2+h ? (I would subtract later time minus earlier time)

Answer 50

yes, and now that we distance moved and the time it took, can you write down the speed (which is distance / time )
?

Answer 51

yes, and now that we have distance moved and the time it took, can you write down the speed (which is distance / time ) ?

Answer 52

19.6h+19.6h^2/h

Answer 53

yes,
can you simplify that ?

Answer 54

\[ \frac{19.6h + 19.6h^2}{h} = \frac{19.6h}{h} + \frac{19.6h^2}{h} \]

Answer 55

19.6h^2+19.6

Answer 56

almost

Answer 57

oh nvm

Answer 58

we are almost done.

Answer 59

the "h" divides into both terms up top
you get
19.6 + 19.6 h

Answer 60

hahahah

Answer 61

the last step, is let h get *very* small. (in other words we want to find the speed when t and t+h are *very close*

we write it as
\[ \lim_{h \rightarrow 0} 19.6 + 19.6 h \]

Answer 62

okk

Answer 63

any idea what the answer is ? (thought not exactly correct, people pretend h is 0 and simplify)
what do you get ?

Answer 64

not sure

Answer 65

\[ \lim_{h \rightarrow 0} 19.6 + 19.6 h \]
when h is zero what is 19.6 + 19.6*h ?

Answer 66

19.6

Answer 67

that is the speed at t=2

Answer 68

so thats it? the velocity is 19.6?

Answer 69

The fast way is to use calculus
\[ \frac{ds}{dt} = \lim_{h \rightarrow 0} \frac{s(t+h) - s(t)}{h} \]

if you know how to take the derivative (using the power rule) we would find
\[ \frac{d}{dt} 4.9 t^2 + 350 = 2\cdot 4.9 t = 9.8 t \]
and at t=2, we would get
\[ \frac{ds}{dt} = 9.8\cdot 2 = 19.6\ m/s \]

Answer 70

^that is all the work for calculus? that's so much simpler haha

Answer 71

yes, which is why we learn calculus.

But the reason it works is all the messy stuff we went through up above.

Answer 72

thank you!