Question

a car traveling at 27m/s uniformly accelerates to rest over distance of 60.8m. What is the deceleration rate? How long did it take to stop?

Answer 1

Here are the equations we should use, they are commonly known as "Kinematic Equations"

Answer 2

oh... it got cut off ;-;
http://www.physicsclassroom.com/Class/1DKin/U1L6a1.gif

Answer 3

It's ok. I have those.

Answer 4

First step is to list the information we do have
so

\(\large v_i = 27 \frac{m}{s}\)
\(\large v_f = 0 \frac{m}{s}\)
\(d = 60.8m\)

and we have two unknowns
a = ?
t = ?

make sense so far? :)

Answer 5

Yes.

Answer 6

Hold on.

Answer 7

Good
then to find \(a\)
we look for the equation that doesn't use \(t\)

Answer 8

Thts what it is. Ok and we'll use the top right correct?

Answer 9

Like I don't get wht I'm supposed to fill in the bottom blanks(graph)

Answer 10

so for each problem you fill out the information we know (like I did above) and use the equations given at the top to find the missing variables

Answer 11

Okay. Can we work the first one together so I kinda know how to do the rest?

Answer 12

A car uniformly accelerates from rest to a final speed of 88 mph (39.2 m/s) in 4 seconds.
What is the acceleration rate of the car? How far did it travel?

^ this one right?

Answer 13

Yes.

Answer 14

I had 78.4 in the first box then
v(m/s) = 39.2
a(m/s^2) = 9.8
t(sec) = 4
I don't know the answer for
V0(m/s)

Answer 15

\(v_0\) means the starting speed
for this, the car starts "from rest"
which means that it's starting speed is 0 m/s

Answer 16

So 0?

Answer 17

yes :)

Answer 18

Okay. Now the second one lol

Answer 19

A car traveling at 27 m/s uniformly accelerates to rest over a distance of 60.8m. What is the
deceleration rate? How long did it take to stop?

so the starting speed is 27 m/s
the ending speed is 0 (because it "accelerates to rest")
distance is 60.8
and we would have to solve for the acceleration and time

Answer 20

So 60.8 goes in the first box right?

Answer 21

yes :)

Answer 22

27 goes in the second?

Answer 23

correct ^_^

Answer 24

Okay so wht would I put down for the v?

Answer 25

it says "accelerates to rest"
"rest" means a speed/velocity of 0

Answer 26

gotcha

Answer 27

Okay this part confuses me cuz idk which formulas to use for which questions

Answer 28

We have two missing things
acceleration and time
if we want to find acceleration, we use the equation that is missing time
if we want to find time, we use the equation that is missing acceleration

Answer 29

I still don't get it. All the problems have either one or the other...

Answer 30

Let's try finding acceleration first together then? :)

we would use the equation that is missing time

Answer 31

Okay

Answer 32

\((v)^2 = (v_0)^2 + 2ad\)

this equation ^_^

Answer 33

then we plug in what we already know

Answer 34

so plug in and what would we get? :)

Answer 35

0^2= 27^2+2(idk)(60.8)
Is tht right?

Answer 36

yes :)

then we try to solve for (idk)
like in algebra

Answer 37

Hahha ok lol

Answer 38

I got 850.6

Answer 39

ehhh can't do it like that

think of it like this
0= 27^2+2(60.8)x

and "solve for x"

Answer 40

Im getting the same thing

Answer 41

0 = 27^2+2(60.8)x
-(27^2) -(27^2) minus (27^2) from both sides
-----------------------------------
-(27^2) = (2*60.8)x
/(2*60.8) /(2*60.8) divide both sides by (2*60.8)
-----------------------------------
\( \large x = \frac{-(27^2)}{2*60.8}\)

Answer 42

Ohhhhhhhh ok hold up

Answer 43

I was using a calculator

Answer 44

you can't use calculators when there are variables

Answer 45

Yea ik. i wanted to get this over with lol

Answer 46

that ^ would get you your acceleration

Answer 47

So -6.0?
I rounded it off from -5.99

Answer 48

yup ^_^

Answer 49

Okay so now time. Would I use this?:
v=v0+at?

Answer 50

use the one without acceleration (because that was rounded and inaccurate)

d = (1/2)*(v + v_0)*t

Answer 51

Okay

Answer 52

\( \large d = \frac{v+v_0}{2} * t \)

d = 60.8
v = 0
v0 = 27
and we solve for t

Answer 53

0+27/2 *T
13.5?

Answer 54

60.8 = 13.5 t

now divide both sides by 13.5

Answer 55

4.5

Answer 56

there you go :)

Answer 57

Okay. I'm still iffy on these problems lol

Answer 58

Can we do a few more if you don't mind?

Answer 59

Try some yourself, it's basically Algebra 1 problems...

Answer 60

Ok

Answer 61

problems like these hide pretty well as math problems (so you can ask it in the math section) ^_^"

Answer 62

Okay. Thanks.

Answer 63

I got 5.76 for number 3. Is that right?