Question

Kinematics,


Who's gonna help me with that cause im going crazy ? :(
https://www.dropbox.com/s/a5f7u18f238rcye/Screenshot%202015-12-15%2013.33.09.png?dl=0

https://www.dropbox.com/s/0esiko2rhid7c75/Screenshot%202015-12-15%2013.34.18.png?dl=0

Answer 1

part (i) is solved btw , i am stuck afterwards , I have also posted the link to the graph (second link)

Answer 2

DO I SIGN in or i don't ?

Answer 3

not

Answer 4

ok thanks

Answer 5

I mean you don't have to sign in on dropbox to view the files

Answer 6

well witch one is better

Answer 7

?

Answer 8

is it better to sign in or not

Answer 9

nvm

Answer 10

Are you familiar with derivatives?

Else I'd suggest you use the utility of excel that calculates it (if it has any).

Answer 11

i am familiar

Answer 12

Well, I recommend you derivate the given function \(x(t)\) and then plot in the point t=1.6.

Answer 13

yeah
find the 1st derivative of x(t) and put t=1.6 in the obtained expression
the value u'll get will be the slope of line tangent to the point on the graph with t=1.6

Answer 14

if I find the first derivative that would give me the velocity while the exercise asks for the aceleration

Answer 15

and if I take the second derivative, then there is nowhere for me to plug in 1,6

Answer 16

so there is that :D

Answer 17

acceleration is \(a=\frac{ dv }{ dt }\) this implies that you'll have to derive twice.

Answer 18

which gives me a(t) = 8

Answer 19

Well done, do not forget the units of measurement.

Answer 20

@ParthKohli could you help me out ? (:

Answer 21

Yes sure.
But I have no idea about Excel.

Answer 22

I have done the excell part
https://www.dropbox.com/s/35sv5hgup6vv4fy/Screenshot%202015-12-15%2014.20.25.png?dl=0

Answer 23

Yeah lol I'd rather use Wolfram.

Answer 24

oki

Answer 25

ok what do you need help with

Answer 26

I dont know how to (b)

Answer 27

even what it's asking seems wrong to me "use x(t) to graph tangent and find accel"

Answer 28

Yes, exactly. You can't find the acceleration just by looking at one tangent. You can only find the velocity through that.

However, if you look at a series of tangents, then you will see a very interesting trend of their slopes and can find the acceleration.

Answer 29

hmmm what do you mean ??

Answer 30

OK, I just meant to say that you can only gather info about velocity by a tangent at a point.

Answer 31

what if I find the slope of the second graph I drew in excell

Answer 32

you'll see that direction of change in velocity will be the direction of acceleration

Answer 33

even then a = -10 while if I calculate it using the expression that he gave me (which I used to contruct the graphs) I get a = 8

Answer 34

not if the graph is of velocity against time
tehn the tangent at a point is the acceleration at that point

it is th esame as the second derivative of the displacement graph

Answer 35

yes, you'll need to first graph v-t

Answer 36

yeah

Answer 37

second graph is v(t) https://www.dropbox.com/s/v6wfgjn3tnd1ndy/Screenshot%202015-12-15%2014.27.45.png?dl=0

Answer 38

slope = -10

Answer 39

x(t)'' = a(t) = 8

Answer 40

so graph shows -10 , expression shows 8 for any value plugged in

Answer 41

so it seems that I did a mild mistake in calculating the slope. The actual slope = -8 . But still , you get a(t) = 8 for any value plugged in (which is diferent sign )

Answer 42

it looks to me like oyu have the slope for the SECOND graph wrong

(there are no titles on the graph)

If the first graph is displacement against time then it can be seen that teh slope (velocity) is DECREASING with time

Hence teh velocity slope would be -8t

and the acceleration would be -8