Question

@zepdrix The displacement in meters of an object moving in a straight line is given by s=1+2t+(1/4)t^2 where t is measured in seconds. (i) [1,3] (ii) [1,2] (iii) [1,1.5] (iv) [1,1.1]
b) find the instantaneous velocity when t=1

Answer 1

i use the formula lim h->0 f(a+h)-f(a)/h right?

Answer 2

Ummm without the limit, yes. The limit tells us the "instantaneous rate of change", the same as the derivative.
In this problem they want the "average rate of change", so we'll just use the difference quotient (the thingy in your limit problem without the limit attached to it).

Answer 3

ok then that is 1/2a+1/4h

Answer 4

Umm it's a little confusing the way they wrote it all abbreviated like that...
I assume (i) is asking for the "average velocity from 1 to 3 seconds".

Answer 5

I guess. it confused me

Answer 6

\[\large v_{ave}=\frac{ s(3)-s(1) }{ 3-1 }\]
This is how we'll find the average velocity between 3 and 1.
I'm not sure what I was rambling on about with that whole limit thing XD lol

Answer 7

Maybe you remember this idea from algebra.
\[\large m=\frac{ y_2-y_1 }{ x_2-x_1 }\]
We're doing the same thing with our problem, but we're using the fancy function notation and letting x and y be time and displacement.

Answer 8

oh so its rate of change? then why does it say displacement? that is the same thing right?

Answer 9

We're taking the idea of rate of change, and giving it a direct application. It's important because displacement, velocity, acceleration have a very clear connection to one another so it's important to get comfortable with the notation. Especially if you ever plan on taking a physics course. :)

Answer 10

Do you understand how to calculate s(3) and s(1)? :)

Answer 11

not really

Answer 12

its not just 3-1 is it?

Answer 13

\[\large s(t)=1+2t+\frac{ 1 }{ 4 }t^2\]
\[s(3)=1+2(3)+\frac{ 1 }{ 4 }(3)^2\]
Understand? :)
Try to find s(3) and s(1).

Answer 14

hmmm I think I tried that once and thought it was wrong

Answer 15

Hmmm :o

Answer 16

well for s(3) I got 37/4

Answer 17

s(1)=13/4

Answer 18

k looks good.
plug them bad boys in!

Answer 19

wait what do I do with those afterwards? I do that for all the intervals?

Answer 20

\[\large v_{ave}=\frac{ s(3)-s(1) }{ 3-1 }\]
You throw them into this formula to find the average velocity between time 1 and 3 seconds.

Answer 21

crap right.

Answer 22

so 37/4-13/4/37/4-13/4 ?

Answer 23

the bottom is just 3 and 1.

Answer 24

so I plugged them in and the velocity appears to be decreasing

Answer 25

\[\large \frac{ s(3)-s(1) }{ 3-1 }=\frac{ \frac{ 37 }{ 4 }-\frac{ 13 }{ 4 } }{ 3-1 }=\frac{ \frac{ 24 }{ 4 } }{ 2 }=\frac{ 24 }{ 8 }=3\]
Hmm did you come up with something like this?

Answer 26

yup then 1, 42/16,1010/400

Answer 27

Oh you did the others as well? neato.

Answer 28

so what do I do with them? where does instantaneous velocity fit it? when I take the limit as x approaches 1?

Answer 29

\[\large s'(t)=v(t)\]v(1)=instantaneous rate of change of velocity at t=1.

Answer 30

so I take the derivative?

Answer 31

mhm

Answer 32

you know Im very tired. I cant think right now. Im going to tackle the rest of this tomorrow. thanks for all your help.

Answer 33

hehe ok np c: haf good nite