How Long to Accelerate for Desired Time?

[pointintime]

Homework Statement

A runner hopes to complet the 10,000 m run in less than 30.0 min. After running at constant speed for exactly 27.0 min, there are still 1100 m to go. The runner must tehn acceelerate at .20 s^2 m for how many seconds in order to achieve teh desired time?

Homework Equations

def of average velocity = t^-1 (X - Xo)

def of a = t^-1 (V - Vo)

Def of average velocity (when a is zero) = 2^-1 (V + Vo)

V = Vo + at

X = Xo + Vo t + 2^-1 a t^2

V^2 = Vo^2 + 2a(X - Xo)

The Attempt at a Solution

ok for the first part of the race I calculated his velocity has to be

5.494 s^-1 m

ok in order to calculate the time to accelerate for the last part of the race

I saw that

X = Xo + Vo t + 2^-1 a t^2

that would do me no good because I do not know the X value for we could stop accelerating and there are two t variables

V^2 = Vo^2 + 2a(X - Xo)

doesn't work becasue we don't know the V^2 at the end of the acceleration

def of a = t^-1 (V - Vo)

rearanged for t didn't work because we don't know the V at the end of the acceleration...

So I'm lost...

 

[kuruman]

You need to figure out the symbols that correspond to the numbers you have.

Forget the initial 27 minutes of the race.

What counts here is that he has 3 minutes left to cover 1100 meters, when his initial velocity is 5.494 m/s. So you have a displacement (1100 m), a time (3 min or 180 seconds) and an initial velocity (5.494 m/s). You need to find an acceleration. What equation do you think you should use?

By the way, when the acceleration is zero, the velocity does not change so the average velocity, the instantaneous velocity and the initial velocity are all the same, v₀.

 

[pointintime]

The question gives us his acceleration for the second part of the race

The runner must tehn acceelerate at .20 s^2 m for how many seconds in order to achieve teh desired time?

were asked for time requirered for him to accelerate at the at that speed in order to achieve that desired time

 

[pointintime]

If he accelerates at .20 s^2 for the rest 180 seconds he would travel

13130 m total which is to long

i believe the problem is asking how long must he accelerate at .20 s^2 then stop accelerating and remain at that velocity for the rest of the race...

Maybe I read it wrong... It's problem 46 in my book so the answer is not in the back of my book and it's a level (III) problem out of 3 levels...

 

[pointintime]

30.0 min = 1800 s
27.0 min = 1620 s

 

[kuruman]

.

Maybe I read it wrong... It's problem 46 in my book so the answer is not in the back of my book and it's a level (III) problem out of 3 levels...

You didn't read it wrong, I did. Your interpretation is correct. Let's use some symbols.
Let v₀=5.494 m/s

Then the distance he travels while he is accelerating (call it x₁) is

x₁=v₀t₁+(1/2)at₁²

The time that you want is t₁ but you need one more equation to find it because you don't know x₁. So note that his final speed after he is done accelerating is

v_f=v₀+at₁

Then he travels the rest of the distance (1100 m - x₁) in time t₂. Then
1100 m - x₁ = v_f t₂ = (v₀+at₁)t₁

Do you see how to finish the solution now?

 

[pointintime]

hmmmm let me see

How do you derive this

1100 m - x1 = vf t2 = (v0+at1)t1

ok I agree with this part...
1100 m - x1 = vf t2
because there is no acceleration after he stopped acceleration so you can just do
Vo t to find the X but why do you set it equal to (v0+at1)t1 ?

vf=v0+at1
I know were that equation comes from the defintion of acceleration

I also know how to derive this
X = Xo + Vo t + 2^-1 a t^2

I'm just unsure were the equation you gave me came from... Could you please tell me how you got it?

THANKS!

 

[kuruman]

After he is done accelerating, his velocity is constant. Call it v_f. The distance that he has left to cover is the initial 1100 m less the distance he already covered under acceleration, 1100 m - x₁. Since he travels at constant velocity he covers equal distances in equal times, therefore

1100 m - x_f = v_f*t₂. OK so far?

Now use v_f = v₀ + a t₁ to replace v_f in the above equation.

 

[pointintime]

hmmmm ok thanks!

 

[pointintime]

Wait...
Is it this
1100 m - x1 = vf t2 = (v0+at1)t1
or this
1100 m - x1 = vf t2 = (v0+at1)t2

 

[kuruman]

Sorry, the second one

1100 m - x1 = vf t2 = (v0+at1)t2

Too much cutting and pasting ...

 

[pointintime]

wait sorry I'm kinda bad

ok i actually understand that

1100 m - x1 = vf t2 = (v0+at1)t2

but how do I solve for t1 the time interval at which the acceleration occuered because I don't know t2 the time that it took to complete the rest of the run after he stopped accelerating I do not know two of the variables

don't know
X1 = the distance covered after the acceleration occuered to finish the race
vf = the velocity at the end of the acceleration
t2 = the time to complete the rest of the race after accelerating
t1 = how long the acceleration occuered

I guess I don't know how to solve for t1 not knowing those variables

know
a = the acceleration given to us in the question .20 s^-2 m
Vo = the velcoity that had to be achieved when he wasn't accelerating in order to cover the first 8900 m at which no acceleration occuered which I calculated to be 5.494 s^-1 m which would be the initial velocity before accelerating
t1 + t2 = 3.00 min sense that would include the time to accelerate and the time after accelerating sense he started to accelerate after 27.0 min and hopes to get under 30.0 min

 

[kuruman]

But you know that t₁ + t₂ = 3 min. So ...

 

[pointintime]

oh.. duh let's see here
:O
ok I would believe that I would solve for t2
right and get
t2 = 3.00 min - t1

X = Xo + Vt2 = Xo + (Vo + at1)t2
X = Xo + V(3.00 min - t1) = Xo + (Vo + at1)(3.00 min-t1)
:|

 

[pointintime]

solve for t1

X = Xo + Vt2 = Xo + (Vo + a(3.00 min - t2))t2

not sure what to do

 

[kuruman]

Get rid of t₂

t₂ = 3 min - t₁.

Now you know that

1100 m - x₁ = (v₀+at₁)t₂
therefore
1100 m - x₁ = (v₀+at₁)(3 min - t₁)
and you also know that
x₁=v₀t₁ + (1/2)at₁²

Take this value for x₁, put it in the previous equation and solve for t₁.

 

[pointintime]

1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t1)

OMG

I got down to this and didn't know what to do

Xo + Xo - 1100m = - Vo t - 2^-1 a t^2 - (Vo + at) (3.00 min - t)

 

[kuruman]

There is no x₀; it is zero. Look at my previous posting and do what I suggested. Do the algebra carefully. You should end up with an equation where t₁ is the only unknown quantity.

 

[pointintime]

ok but is this step right

1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t1)

were I just plugged in X

then take out the Xo s
is that correct and then just solve for t?

 

[kuruman]

Your equation is sloppy. It mixes t and t₁. There is only one kind of time, and that's t₁. Also you show (v₀ at). It should be (v₀+at₁). So fix it up and solve for t₁.

 

[pointintime]

1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t)
is there any other subsitution or is just solve for t?

 

[kuruman]

Please read posting #20 and fix your equation first.

 

[pointintime]

done to this

(3.00 min - t)^-1 (1100 m - Xo - Xo - Vo) = 2^-1 a^2 t^3 + Vo t

don't know were to go from here

 

[pointintime]

1100 m - (Xo + Vo t + 2^-1 a t^2) = Xo + (Vo at)(3 min - t)
I thought they were all t1 were every single t is equal to t1

 

[pointintime]

1100 m - x1 = (v0+at1)(3 min - t1)

subsitute in this
x1=v0t1 + (1/2)at12

gave me this did I do it wrong?

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

 

[pointintime]

t^3 = (3.00 min - t + a^2)^-1 2(1100 m - 2 Xo - Vo t - Vo)

I was going to take out the Xo s at the end

 

[pointintime]

rewrote for zero

0 = (2(1100 m - 2 Xo - Vo t - Vo))^-1 t^3(3.00 min - t + a^2)

 

[pointintime]

ok well i think i have done it wrong i believe this part however is correct

1100 m - 2Xo = (Vo + at)(3.00 min - t) + Vo t + 2^-1 a t^2

 

[pointintime]

anyone might know how to do this?

 

[pointintime]

I don't know if I can get any closer wihout pluggining in somehting which i don't know what

(1100 m - 2 Xo - Vo t - Vo) (a(3.00 min - t))^-1 = t + 2^-1 t^2

 

[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 occured

 

[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]

Dosen'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 dosen'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 occured

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]

?