How do I use physics equations for this?

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A vehicle accelerates from rest at 2.0 m/s² until reaching 20 m/s, taking 10 seconds, then decelerates at 1.0 m/s² until it stops, taking an additional 20 seconds. The total time from start to stop is 30 seconds. The total distance traveled during acceleration is 100 meters, and during deceleration, it is calculated to be 300 meters, resulting in a total distance of 400 meters. The discussion highlights the importance of using appropriate physics equations for solving kinematics problems. Understanding the distinction between speed, velocity, and acceleration is crucial for applying the correct formulas.
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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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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:
 
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 :)
 
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
 
Hmmm I seem to have my vocabulary messed up...
Would it be correct to say
a = distance/time?
 
Last edited:
(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
 
So what exactly is the rate that distance divided by time gives? Is it not acceleration?
 
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:​
 
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:
 
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