Question

https://csusb.blackboard.com/bbcswebdav/pid-351838-dt-content-rid-1406724_1/courses/148PHYS12101/practice%20mid%20term%20121%20fall%202014%281%29.pdf

Answer 1

If someone could help me with number 3 that would be greatly appreciated.

Answer 2

need log in information
send me pm if I am not here , send message to inbox, i will check and help you

Answer 3

will definitely help

Answer 4

alright

Answer 5

1. g = 9.8 m/s^2

Answer 6

Got that one.

Answer 7

lets start with the first word problem

Answer 8

Is the first step to convert the 4m/s^2 to standard si units?

Answer 9

@perl

Answer 10

yes one sec

Answer 11

Take your time!

Answer 12

ok

Answer 13

i am integrating , have you done some calculus?

Answer 14

I sorta get whats goin on here

Answer 15

I have but im bad at it. Wont I just sqroot the 4 to get 2m/s then divide and get 10?

Answer 16

car equation:
a= 0
v= 20
x = 20t +0 (since he starts from origin)

police officer.
a = 4
v = 4t + 0 (since he starts from rest)
x = 4*t^2/2 + 0 (since he starts from origin)

Answer 17

20t = 4*t^2/2

20t = 2t^2

2t^2 - 20t = 0

2t ( t - 10) = 0

t = 0, 10

Answer 18

so t=10

Answer 19

This is integration you said?

Answer 20

yes, or you can use the kinematic equations (where we assume constant acceleration)

Answer 21

Now we're talking about stuff I sorta understand

Answer 22

v = da/dt
x = dv/dt

Answer 23

ok :)

Answer 24

ill keep calculus out then.
there should be 3 or 4 kinematic equations in your notes

Answer 25

If we can keep the calculus out of this as much as possible that would be better. My professor doesnt even like to use it.

Answer 26

there are

Answer 27

the constant acc one and the free fall one

Answer 28

ok . we can use the one which says x in it

Answer 29

x = 1/2 * a t^2 + v0*t + x0

Answer 30

yes

Answer 31

so for the car, constant velocity implies acceleration = 0. That leaves us with
x = v0*t + x0

Answer 32

x=vot+vo+1/2at^2

Answer 33

you have a typo there

Answer 34

x = x0 + v0*t + 1/2*a*t^2

Answer 35

oh i see. Now quick question

Answer 36

can there ever be a - sign in front of the 1/2?

Answer 37

theres one that deals with that later on.

Answer 38

yes, if acceleration is negative. in free fall acceleration points down, so we have negative in front of g.

Answer 39

ok thanks I was getting thrown off by that. lets see if I can solve the car problem with the constant acc equation.

Answer 40

but free fall (and projectile motion) is a special case. generally speaking it depends on the problem what is the acceleration

Answer 41

ok.

Answer 42

in free fall situation, usually we just say acceleration = -g , and you don't need a whole separate equation

Answer 43

ok so im kinda lost with using this equation here.

Answer 44

ok, there is a separate equation for the car and the police cruiser

Answer 45

you can also use the fact that

distance = average speed * time

Answer 46

s=v/t

Answer 47

x = <v>*t , that will be for the car , since the velocity is constant.

this equation won't work for the policeman, since he does not have constant velocity (since his acceleration is not zero )

Answer 48

ok.

Answer 49

thats wrong

v = s/t

Answer 50

speed = distance / time , s stands for distance or displacement , v is usually used for velocity (speed with sign)

Answer 51

ok so im trying to figure out what to plug in where.

Answer 52

v= s/t , so s = v*t (this is only valid for constant velocity)

Answer 53

ok so just multiply the speed and time.

Answer 54

v = 20 for the car, so

s = 20*t

Answer 55

ok but theres no time avaliable isnt that what we're trying to find?

Answer 56

first we want to find equation of distance for each guy

Answer 57

then we will set the distance equations equal to each other, to find the time where they meet

Answer 58

ok.

Answer 59

where they meet, distances are equal

Answer 60

let me ask though the way I explained earlier. The way I initially did it. Will that work most times. It was more of a common sense thing I did.

Answer 61

what was the way you initially did it?