How Long to Accelerate for Desired Time?

[kuruman]

Start from

1100 m - [v₀t₁+(1/2)a t₁²]=(v₀+at₁)(3 min - t₁)

and solve for t₁.

 

[pointintime]

1100 m - [v0t1+(1/2)a t12]=(v0-at1)(3 min - t1)
I thought it was
1100 m - [v0t1+(1/2)a t12]=(v0+at1)(3 min - t1)
:O

 

[kuruman]

It is, I corrected it.

 

[Jotun.uu]

It seem's you are having problems solving for t1.

I'll crash through it real quick and see if that helps.

I'm going to make it a bit easier for myself first though:

s = 1100m
v = v0
t = t1
a = a

Remember that if you do something on the "right side" of the "=" you have to do it to the "left side" as well. So adding something to one side is the same as adding it to the other (which can be useful if you for example have t + 2 = 7 --> t + 2 - 2 = 7 - 2 which means t = 5).

new equation: s - (vt + (1/2)at²) = (v+at)(3-t)
- vt - (1/2)at² = 3v - vt + 3at - at² - s
move everything with "t" in it to the left hand side:
- (1/2)at² + at² - vt + vt - 3at = 3v - s
add up the t and t²:
(1/2)at² - 3at = 3v - s
make t² "alone" by multiplying with 2 and dividing both sides with a:
t² - 6t = 6v/a - 2s/a

Now just use a formula and solve for t. And remember that t > 0! Hope I didn't make it too easy for you, and also be sure to figure out what I did while I was solving for t!

 

[pointintime]

1,100 m - [Xo + Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3)
My algebra is kind of weak can I just add the brackets to the left side and then take them off? How doI then go about solving for t3??

1,100 m = (a t3 + Vo) (180 s - t3)+ [Xo + Vo t3 + 2^-1 a t3^2]

?

Wow looks at equation...

 

[pointintime]

Ok I'm here right now and will continue...

((a t3 + Vo)(180 s - t3))^-1 -[Xo + Vo t + 2^-1 a t3^2] = - 1,100 m

do I just multiply this out
((a t3 + Vo)(180 s - t3))^-1

I don't see how anything is suppose to cancel out though...

 

[pointintime]

((a t3 + Vo)(180 s - t3))^-1 = ( a t3 (180 s) - a t3^2 + Vo (180 s) - Vo t3)^-1

What's suppose to cancell out?

 

[pointintime]

I thought you couldn't cancel out inverse and normal numbers if there was addition I thought you could only when there was multiplication...

(AB)^-1 (AC)

you would be able to cancel out A

 

[pointintime]

oh i got it

 

[kuruman]

Nothing is supposed to cancel out. Please follow Jotun.uu's suggestion. The bottom line is a quadratic equation. You need to put in the symbols and solve it.

 

[pointintime]

don't worry still working on it
you didn't make it to easy lol

 

[kuruman]

I admire your persistence. Keep up the good work.

 

[Jotun.uu]

don't worry still working on it
you didn't make it to easy lol

but if you work hard on something you remember it a lot better =)

 

[pointintime]

ok well the 2^-1 is a half and the 3 is actually 3 min

I got down to here

- 2 ^-1 a t3^2 - (180 s)(a t3) + a t3^2 = Vo(180 s) - 1100 m + Xo

and I was about to divide by a when I realized you couldn't...

- 2 ^-1 a t3^2 - (180 s)a + (180 s)t3 + a t3^2 = Vo(180 s) - 1100 m + Xo

see...

(180 s)(a t3)
this became

(180 s)a + (180 s)t3

preventing from dividing by acceleration

also note that instead of s I used 1100 m and instead of 3 I used 180 s for 180 seconds
instead of using 3 or 3 min because so that way I wouldn't have to convert to seconds later and that way I wouldn't confused and just use 3 without any units...

also how come for Xo you said it was zero when we can't use 8900 m the point at which the acceleration occurred

 

[Jotun.uu]

using 180 instead of 3 will only make it so that the answer will be in seconds instead. my final equation would then be t² - 360t = (360v-2s)/a

 

[pointintime]

doesn't this prevent you from dividing by acceleration?

(180 s)(a t3)
this became

(180 s)a + (180 s)t3

preventing from dividing by acceleration

If it doesn't then why not?

 

[Jotun.uu]

180*a*t3 is not equal to 180a + 180t3 [[a*b*c IS NOT a(b+c)]]

in the same way that if
example:
2*3*4 = 24 while 2*3 + 2*4 = 6 + 8 = 14

 

[Jotun.uu]

also how come for Xo you said it was zero when we can't use 8900 m the point at which the acceleration occurred

We don't need Xo in our equations because the distance (8900m) up until the final stretch of 1100m doesn't matter. the only thing that matters is the initial velocity, the remaining distance, the acceleration and the time needed.

 

[pointintime]

It wasn't I believe there were parantheses or however you spell forcing you to distribute correct?

- (180 s)(a t3)

I'll make sure that there were

this was line before

-[Vo t3 + 2^-1 a t3^2] = 180 s (a t3) - a t3^2 + Vo (180 s) - Vo t3 - 1,100 m + Xo

line before that

-[Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3) - 1,100 m + Xo

so there would be no ( ) around a t3?

like aren't you doing this...

(a t3)(180 s)
?

and here's the line before that line

1,100 m - [Xo + Vo t3 + 2^-1 a t3^2] = (a t3 + Vo) (180 s - t3)

 

[pointintime]

?

 

[pointintime]

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

 

[rl.bhat]

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

You can solver this problem very easily by drawing time- velocity graph.
You can find the required time by considering the area above the last time interval i.e. 180 seconds.
This area consists of a rectangle with area 5.494*180 = 988.92 m.
The remaining area = 1100 - 988.92 = 111.1 m.
This distance is equal to the area of the trapezium with parallel sides 180 s and t2 s, and height 0.2*t1 .
So 111.1 = 1/2*0.2*t1(180 + t2) Since t1 + t2 = 180 s, t2 = (180 - t1)
Hence 111.1 = 1/2*0.2*t1(180 + 180 - t1)
Or
111.1 = 1/2*0.2*t1(360 - t1)
Solve this quadratic to find t1.

 

[pointintime]

um 5.494 is the intial velocity of the person running before the acceleration occurs in which an acceleration is present and at the end of 180 second the person is no longer running in fact he stops accelerating before the 180 seconds, the end of the race, the question asked for how much time he msut accelerate in order to get under 30 min

 

[pointintime]

So I don't see how your way works

 

[rl.bhat]

So I don't see how your way works

During 180 seconds he accelerates for t1 seconds and runs uniformly for t2 seconds. Both these are taken into account in the area of the trapezium.

You solve the quadratic for t1 and check the answer.

 

[kuruman]

There is absolutely no reason to have x₀ in your expression.

Step by step help

1. Go back to posting#45 by Jotun.uu.
2. Read it very carefully.
3. In the equation shown in #45, enter the numbers your have for the acceleration a = 0.2 m/s². v = 5.494 m/s and s = 1100 m
4. Solve the quadratic equation for t. If you are not sure how to do this, google "quadratic formula".

 

[pointintime]

Ok I graphed into a graphic calculator

y1 = -2 X^2 - 180 x + X^2

y2 = .2^-1 (5.494*180-1100)

and got this

(3.111, -555.4)

I assume I need the 3.111 seconds for my answer
but sense this is physics not algebra how do I rearange for y3?

 

[pointintime]

wait so I just forget

.2^-1 (5.494*180-1100)

that part of the equation and just do the opposite of b square root... yada yada yada?

 

[kuruman]

The correct answer is 3.11 s. The negative answer is unphysical. I don't know what y3 means.

 

[pointintime]

x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

so is the 2a part...

the a is -2^-1 so does it become part of numerator and I get this?
x = a(-b +/-sqrt(b^2 - 4ac))/2 ("sqrt" = square root)