Question

An electric vehicle starts from rest and accelerates at a rate of 2.6 m/s2 in a straight line until it reaches a speed of 10 m/s. The vehicle then slows at a constant rate of 1.0 m/s2 until it stops.
(a) How much time elapses from start to stop?(b) How far does the vehicle travel from start to stop?

Answer 1

\[\large a_1=2.6 m/s^2\]
\[\large a_2=-1 m/s^2\]
\[\large v_i=0 m/s\]
\[\large v_f=10 m/s\]

this is mostly a substitution problem using your kinematic formulas

Answer 2

\[\large x_f = x_i + v_i t + \frac 1 2 a t^2\]

Answer 3

what is x, if v is velocity, t is time and a is acceleration?

Answer 4

x is position instead of p
p is used to represent linear momentum which you will probably learn about later on

Answer 5

so what numbers or constants can i plug in for x?

Answer 6

ok, so you start at rest correct?

Answer 7

what values can you imply if you start from rest?

Answer 8

and which are most convenient?

Answer 9

yes which equals 0 , right?

Answer 10

remember, whenever possible, you want to use the most convenient values.
this does not necessarily imply 0, but typically does because 0 allows you to cancel out terms

Answer 11

so then what would the value be at rest if it isnt zero because from understanding rest means not moving which would equal 0

Answer 12

okay, I was too vague there

Answer 13

when you are at rest, VELOCITY IS ALWAYS 0
position however is a matter of convenience. typically, if you use 0 that will cancel out the x-initial term, but sometimes maybe you are using a different scale or whatnot
does that make sense?

Answer 14

okay i think i understand but then what kind of factors determine position?

Answer 15

ok, do you know the four basic kinematic formulas?

Answer 16

no

Answer 17

velocity as a function of time
\[\large v_f = v_i + at\]

position as a function of velocity and time
\[\large x_f=x_i + \frac 1 2 (v_i+v_f)t\]

position as a function of time
\[\large x_f = x_i + v_i t+ \frac 1 2 a t^2\]

velocity as a function of position
\[\large v_f^2 = v_i^2 + 2a(x_f-x_i)\]

Answer 18

the important thing to remember though is that all of these equations assume one-dimensional motion

Answer 19

you'll probably get into two dimensional motion later on

Answer 20

as of right now, I would recommend the first equation, since you are looking for time

Answer 21

but remember you need to do it twice, once for acceleration, and once for deacceleration and add the two times

Answer 22

from there, you could probably apply any of the first three to get distance travelled

Answer 23

what do you mean by "and add the two times"

Answer 24

do you mean add the sums together?

Answer 25

you need to find t1 which is when you are accelerating or speeding up
then once you hit v=10m/s, you're going to deaccelerate. But remember than when you deacclerate, v_i and v_f switch

Answer 26

does it make sense? do you understand what you need to do?

Answer 27

okay so i know for time 1 the answer is 10/2.6 seconds but how do i find time 2

Answer 28

ok so now, you're going at v=10 m/s so that is now your initial velocity
you're going to keep going until you stop so final velocity is 0
that's why i said you swap the velocities
your accleration is now -1m/s^2

Answer 29

okay so then the time 2 is 10 s

Answer 30

is it correct?

Answer 31

do you mean the second time in seconds?

Answer 32

no. I don't have the answer. I was wondering if you had the correct answer

Answer 33

\[x _{f}=x _{i}+0+(1/2)(2.6)(10/2.6)^2\]

Answer 34

i still dont know what x is ? i understand its a position but what is its value?

Answer 35

ok, so let's say for a minute that the total distance traveled is 10 units (on an x-y axis) does it matter if you start from -5 and go to 5, or 0 and go to 10? or even start at 52 and go to 62?

Answer 36

no it doesnt matter, 10 units is 10 units

Answer 37

\[\Delta x= x_f-x_i\]
delta x is also distance travelled
so if you don't care where you start, if you move x_i to the left of the equal sign, do you agree that you get the distance traveled?

Answer 38

do you mean \[\Delta(x)-x _{i}=x _{f}\]

Answer 39

\[\large x_f=x_i +v_i t + \frac 1 2 a t^2\]
\[\large x_f - x_i =x_i -x_i+v_i t + \frac 1 2 a t^2\]
\[\large \Delta x=v_i t + \frac 1 2 a t^2\]

Answer 40

this make sense?

Answer 41

yes , so now i have to find what delta x is

Answer 42

no you don't, you already found delta x

Answer 43

look at your earlier post

Answer 44

delta x is 10?

Answer 45

actually i used 10 as an example, I meant YOUR post, not mine

Answer 46

the one where you substituted numbers

Answer 47

okay so then delta x =19.23, is that my answer?

Answer 48

first of all was your total time 13.8s? assuming my approach was correct, that was my answer

Answer 49

yes my total time was 13.8. was i supposed to use the total time to complete the equation because i only used the time for acceleration at 2.6 m/s

Answer 50

I was just checking
but ok yes delta x 1 is 19.2 m

Answer 51

now think about this for a second, earlier you said that 10 units is 10 units no matter where you start. so which location is the most convenient initial position such that the final position IS delta x?

Answer 52

initial position could be zero

Answer 53

remember, in kinematics, these equations are a matter of convenience, you can start where you want but some locationns will make your life hell

Answer 54

YES!

Answer 55

so then 19.2 =\[x _{f}-0\]

Answer 56

making \[x _{f}=0\]

Answer 57

correction x_i=0, not x_f

Answer 58

also remember that you can only do this when you start at rest

Answer 59

i mean \[x _{f}=19.2\]

Answer 60

so for the second part when you deaccelerate your initial position is not 0, it's 19.2

Answer 61

make sense?

Answer 62

yes because you are leaving from the postiion you went to

Answer 63

just as a side note, in THIS senario, there is no difference, but you won't always be that lucky

so if you make the mistake of assuming that your initial velocity is 0 and proceed, you may get lucky, but sooner or later, a senario will come up in which it comes back to bite you

Answer 64

over time you become more experienced and can tell when it doesn't matter

Answer 65

so how do i know what the initital velocity is if its not zero?

Answer 66

it's 10

Answer 67

it's the velocity you were at when you started to deaccelerate

Answer 68

oh okay so do i have to use the formula \[v _{f}=v _{i}+at\]

Answer 69

remember to focus on what you're looking for now
you're looking for distance 2

Answer 70

which is delta x, right?

Answer 71

also remember that you are looking for x_final
x-initial is 19.2 so you can just leave x_final on the left and x-initial on the right

Answer 72

\[x _{f}=19.2+(10)(13.8)(1/2)a(13.8)^2\]

Answer 73

let's call your starting point A
let's call B the point where your velocity hit 10m and your acceleration changes to -1m/s^2
lets call C the point you finally stop

Time time it took to go from point B to point C was 10 s whereas 13.8 was the time it too to go from A to C

based on this, which time should you use?
remember, your initial position right now is 19.2, which is where you stopped when going from A to B and where B starts

Answer 74

for going from B to C

Answer 75

so then the time is 10 s and the acceleration is -1 m/s

Answer 76

yes

Answer 77

then the equation is \[x _{f}=19.2+(10)(10)+(1/2)(-1)(10)^2\]

Answer 78

that's what i got

Answer 79

the answer is 69.2?

Answer 80

well that's what I got
one sec
@LastDayWork
could you do me a favor and check whatI did?
relativity's burned me out so maybe you'll catch something I did wrong

Answer 81

you are right, i checked on my homework . thank you so much !!!!!

Answer 82

well okay then
and you're welcome

@LastDayWork thanks

Answer 83

:)