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#### markosheehan

##### Member

- Jun 6, 2016

- 136

ive set up equations but i am lost in all of the variables

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- Thread starter
- #1

- Jun 6, 2016

- 136

ive set up equations but i am lost in all of the variables

- Jan 30, 2012

- 2,533

Then share them.ive set up equations

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- #3

- Jun 6, 2016

- 136

So v=0+at1 so t1 =v/a doing the same for time 2 you get t2= v/3a

Then using area of a triangle .5*v(v/a +v/3a)

Then I put this into my formula above aswell as the total time and equal it to ✓s/2 and I have an expression for s.

Anyway when I solve this I got no where.

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- #4

- Jun 6, 2016

- 136

any helping info at all?

- Jan 30, 2012

- 2,533

So $t_1=\dfrac{v}{a}$, $t_2=\dfrac{v}{3a}$; therefore, $t_1+t_2=\dfrac{4v}{3a}$. Also, $s=\dfrac{1}{2}v(t_1+t_2)=\dfrac{2v^2}{3a}$.So v=0+at1 so t1 =v/a doing the same for time 2 you get t2= v/3a

Then using area of a triangle .5*v(v/a +v/3a)

The average speed is $\dfrac{s}{t_1+t_2}$, which equals $\sqrt{\dfrac{s}{2}}$ by assumption. Taking squares of both sides, we get $\dfrac{s^2}{(t_1+t_2)^2}=\dfrac{s}{2}$, from where $2s=(t_1+t_2)^2$. Now substitute the expressions for $s$ and $t_1+t_2$. Then $v^2$ should cancel and you get an equation in $a$.

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- #6

- Jun 6, 2016

- 136

Im sorry im not sure about some of your answer.Sorry for the delay.

So $t_1=\dfrac{v}{a}$, $t_2=\dfrac{v}{3a}$; therefore, $t_1+t_2=\dfrac{4v}{3a}$. Also, $s=\dfrac{1}{2}v(t_1+t_2)=\dfrac{2v^2}{3a}$.

The average speed is $\dfrac{s}{t_1+t_2}$, which equals $\sqrt{\dfrac{s}{2}}$ by assumption. Taking squares of both sides, we get $\dfrac{s^2}{(t_1+t_2)^2}=\dfrac{s}{2}$, from where $2s=(t_1+t_2)^2$. Now substitute the expressions for $s$ and $t_1+t_2$. Then $v^2$ should cancel and you get an equation in $a$.

if t1+t2=2v^2/3a and s=2v^2/3a im not sure how 2s=(t1+t2)^2 as 2(2v^2/3a) does not equal (2v^2/3a)^2

im getting a=16/3 when i sub those in thanks.

- Jan 30, 2012

- 2,533

I wrote a different formula.if t1+t2=2v^2/3a