Remember that given a position function we can find the velocity by evaluating the derivative of the position function with respect to time at the particular

?423 2 tvst

by given isvelocity its if at object an of position the is What

1 Example

Remember that given a position function we can find the velocity by evaluating the derivative of the position function with respect to time at the

particular time.

dt

xdtvtx

then , Given

Hence, what we need is an antiderivative of the velocity function.

dtvx

dttx 42 2

ttx 43

2 3

?423 2 tvst


1 Example

ttx 43

2 3

To check our answer we need only take the derivative of the result.

dt

xdv

tt

dt

dv 4

3

2 3

42 2 tv

It’s appears we have found the right function …

or have we????

?423 2 tvst


1 Example

ttx 43

2 3

What if the position function is

543

2 3 ttdt

dv

42 2 tv

543

2 3 ttx

So this function also works!

?423 2 tvst


1 Example

ttx 43

2 3

What if the position function is

1043

2 3 ttdt

dv

42 2 tv

1043

2 3 ttx

So this function works as well!

?423 2 tvst


1 Example

constant re whe CCttx 43

2 3Since the derivative of a constant is 0, it seems that any function of the form

works just as well!

constantany is where

Integral Indefinite

CCxFdxxf

?423 2 tvst


1 Example

msx Suppose 50 In order to find a particular solution you must be given an initial condition.

Cmx 0403

25)0( 3

mC 5

543

2 3 ttx

Particular Solution

53433

23 3 sx

msx 353

Cttx 43

2 3

? if

by given is onaccelerati whose at object an ofvelocity the is What

2 Example

smsvt

smt

sma

st

.90466

2

243

dtav

dtttv 266

Cttv 32 23

Csmsv 32 4243.904

But

smC .10

.1023 32 ttv

? if

by given is onaccelerati whose at object an ofvelocity the is What

2 Example

smsvt

smt

sma

st

.90466

2

243

.1023 32 ttv

.1022232 32 v

smv 142

.000.20.30.2 2 mxtttv and by given is particle a ofvelocity The

3 Example

? at position sparticle' the is Whata. st 0.3

dtvx

dtttx 232 2

Ctttx 22

3

3

2 23

Cx 0202

30

3

200 23

But

0C

tttx 22

3

3

2 23


3 Example

? at position sparticle' the is Whata. st 0.3

tttx 22

3

3

2 23

0.320.32

30.3

3

20.3 23 x

mx 5.10.3


3 Example

? at onaccelerati sparticle' the is Whatb. st 0.3

dt

vda

232 2 ttdt

da

34 ta

30.340.3 a

20.90.3s

ma


3 Example

? to from ntdisplaceme sparticle' the is Whatc. stst 0.40.2

42 dtvx

42

2 232 dtttx

4

2

23 22

3

3

2

tttx

222

2

32

3

2424

2

34

3

2 2323x

3

24

3

210x

mx 15

? at position its is What. position at speed withupward thrown is ballA

4 Example

1ttxv oo

Earth the of surface the near constant ga

dtgdta

Ctgv

Cgvv o 00

But

ovC

ovtgv 1st Kinematic Equation


4 Example

1ttxv oo

tgvv o 1st Kinematic Equation

dttgvdtv o

Ctgtvx o 2

2

1

Cgvxx oo 202

100

But

oxC

oo xtgtvx 2

2

1


4 Example

1ttxv oo

2

2

1tgtvxx oo

3rd Kinematic Equation

oo xtgtvx 2

2

1

2111 2

1tgtvxtx oo

Documents

Remember that given a position function we can find the velocity by evaluating the derivative of the position function with respect to time at the particular