How do I use physics equations for this?

  • Thread starter netrunnr
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In summary: Thanks for catching that. :)In summary, homework statement:a vehicle starts from rest and accelerates at a rate of 2.0m/s2 in a straight line til it reaches a speed of 20m/s. then it slows to a constant 1.0m/s2 until it stops.
  • #1
netrunnr
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Homework Statement



a vehicle starts from rest and accelerates at a reate of 2.0m/s2 in a straight line til it reaches a speed of 20m/s. then it slows to constant 1.0m/s2 til it stops.
how much time elapses from start to stop?
how far does the vehicle go?


Homework Equations


I have:
a1 = 2.0m/s2
a2 = 1.0m/s2
v1 = 2.0m/2
v2 = 0.0m/2


The Attempt at a Solution



what I did was this:
t1=0s then 2m/s2 * 0s= 0
t1=1s then 2m/s2 * 1s = 2m/s
t1=10s then 2m/s2 * 10s = 20m/s

that gives me 10 seconds later the vehicle gets to a = 20m/s

so now the same thing for slowing down setting my 20m/s to t=0 and working from there
t2=0 then 1m/s2 * 0s = 0
t2=1 then 1m/s2 * 1s = 1m/s
t2=20 then 1m/s2 * 20s = 20m/s

now I have:
total time: t2 + t1 = 20s + 10s = 30s

this is the only equation given that I know how to apply:
total distance: based on a = d/t set to solve for d so d = at
total distance: d1 = 2.0m/s * 10s = 20m
d2 = 1.0m/s * 20s = 20m
20m + 20m = 40m

total time traveled is 30s
total distance traveled is 40m

the problem is I did not use any of the basic equations I was given in my book for this. I am at a loss as to how to match my equations to my work... and seeing as this is the first week of the semester I have a feeling this is going to come back to haunt me drastically in about 4 weeks.

thanks for any advice offered.
 
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  • #2
hi netrunnr! :smile:
netrunnr said:
what I did was this:
t1=0s then 2m/s2 * 0s= 0
t1=1s then 2m/s2 * 1s = 2m/s
t1=10s then 2m/s2 * 10s = 20m/s

that gives me 10 seconds later the vehicle gets to a = 20m/s

so now the same thing for slowing down setting my 20m/s to t=0 and working from there
t2=0 then 1m/s2 * 0s = 0
t2=1 then 1m/s2 * 1s = 1m/s
t2=20 then 1m/s2 * 20s = 20m/s

now I have:
total time: t2 + t1 = 20s + 10s = 30s

yes, that's correct, but using v = vo + at would have been a lot simpler and quicker …

that is what physics equations are for!
this is the only equation given that I know how to apply:
total distance: based on a = d/t set to solve for d so d = at

no, a = d/t is wrong :redface:

look up your constant acceleration equations, and choose one that seems appropriate to find the distance(s) …

show us what you get :smile:
 
  • #3
tiny-tim said:
hi netrunnr! :smile:


yes, that's correct, but using v = vo + at would have been a lot simpler and quicker …

that is what physics equations are for!


no, a = d/t is wrong
opps that is r = d/t my bad... :redface:


okay new take:
v = v0+at

with these values for part I:
v =20m/s ; a = 2m/s2 ; t = unknown
solve for t
20m/s = 0 + 2m/s2*t
20m/s * s2/2m = t
t1 = 10s

values for part II:
v = 10m/s ; a =1m/s2 ; t = unknown
10m/s = 0 + 1m/s2 * t
`0m/s * s2/1m = t
t2 = 20s

total time = t1 + ts = 30s

r = d/t solve for d
2m/s2 = d1/10s
2m/s2/10s = d1
d1=20m


1m/s2 = d2/20s
1m/s2/20s = d2
d2 =10m

total distance = d1 + d2 = 20m + 10m = 30m

let me know what you think and thanks for the help so far :)
 
  • #4
netrunnr said:
total time = t1 + ts = 30s

ok so far :smile:
r = d/t solve for d
2m/s2 = d1/10s
2m/s2/10s = d1
d1=20m


1m/s2 = d2/20s
1m/s2/20s = d2
d2 =10m

i don't understand any of this :redface:

is r meant to be speed? (if so, please use v, like everyone else :wink:)

you've written r = d/t (instead of a), but you've used the acceleration anyway
 
  • #5
Hmmm I seem to have my vocabulary messed up...
Would it be correct to say
a = distance/time?
 
Last edited:
  • #6
(just got up :zzz: …)
netrunnr said:
Hmmm I seem to have my vocabulary messed up...
Would it be correct to say
a = distance/time?

a is for acceleration
 
  • #7
So what exactly is the rate that distance divided by time gives? Is it not acceleration?
 
  • #8
speed! :smile:

that's why speed is measured in m/s ! :rolleyes:

(and of course, only if the acceleration is zero)

get some sleep! :zzz:​
 
  • #9
omg a *duh* moment indeed.
lesson here : do not overthink a solution :biggrin:
 
  • #10
just learned something very important today regarding this - and it was the source of my confusion here:
Speed is the rate of change of distance with time.
Velocity is the rate of change of displacement with time.

I thought speed was acceleration... so on that note:

my problem is a total time of 30s
and the distance traveled is:
x = xo + vot = 1/2at2
so part one is x1=0 + 0 + 1/2*2*102 = 100m
then part 2 is xtotal=100 + 20m/s+1/2*(-1)*202 = 300m

so the total distance traveled is 300m! not 30.

is this right?
 
  • #11
hi netrunnr! :smile:
netrunnr said:
I thought speed was acceleration...

ah, that explains a lot!

i was getting worried! :biggrin:
my problem is a total time of 30s
and the distance traveled is:
x = xo + vot = 1/2at2
so part one is x1=0 + 0 + 1/2*2*102 = 100m
then part 2 is xtotal=100 + 20m/s+1/2*(-1)*202 = 300m

yes, that's fine! :smile:

(warning: in an exam, don't do two lines in one, as in your last line …

there's nothing wrong with it, but there's so much chance of making a silly mistake that it's not worth it :wink:)

btw, you can also get there without using t, by using v2 = u2 + 2as (twice) …

try it! :smile:
 

1. How do I determine which physics equations to use for a specific problem?

First, identify what type of problem you are solving. Is it a motion problem, a force problem, or an energy problem? Then, refer to your physics formula sheet or textbook to find the appropriate equations for that type of problem.

2. Can I use multiple equations in one problem?

Yes, you can use multiple equations in one problem as long as they are applicable and you are solving for the same variables. Make sure to properly rearrange and substitute the equations to avoid errors.

3. How do I know which variables to solve for in a physics equation?

Look at the given information in the problem and determine which variables are known and which ones are unknown. The unknown variables are the ones you need to solve for using the appropriate equations.

4. Can I use the same physics equation for different types of problems?

No, each physics equation is specific to a certain type of problem. For example, you cannot use an equation for motion to solve a problem about forces. Make sure to use the correct equation for the specific problem you are solving.

5. How do I check if my answer is correct when using a physics equation?

You can check your answer by substituting it back into the original equation and making sure it satisfies the given information in the problem. You can also use unit analysis to check if your answer has the correct units.

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