Prelim 2016 Maths Help Thread (3 Viewers)

leehuan · May 16, 2016

eyeseeyou said:
You got a right but for b it was a=-1 and for some reason I can't find C there

Of course acceleration = -1. Did you sub t=0 in?
______________

To find C, use what the question said.

The question explicitly states when t=0, x=0

eyeseeyou · May 16, 2016

leehuan said:
Of course acceleration = -1. Did you sub t=0 in?
______________

To find C, use what the question said.

The question explicitly states when t=0, x=0

ty for your help leehuan

InteGrand · May 16, 2016

eyeseeyou said:
yeah but the parabola isn'tsupposed to cross into the negative side

Why not?

davidgoes4wce · May 16, 2016

I thought the parabola could cross the negative. Didn't mention anywhere in the question that the distance was 'always to the right'.

eyeseeyou · May 16, 2016

InteGrand said:
Why not?

And now Integrand comes to the rescue after so long...lol

because the answer sheet shows that it doesn't cross into the negative side

eyeseeyou · May 16, 2016

eyeseeyou said:
And now Integrand comes to the rescue after so long...lol

because the answer sheet shows that it doesn't cross into the negative side

But I think it can...I'm pretty sure it can

InteGrand · May 16, 2016

eyeseeyou said:
And now Integrand comes to the rescue after so long...lol

because the answer sheet shows that it doesn't cross into the negative side

$\noindent Did the answer sheet give the equation for $x(t)$? The one davidgoes4wce has is correct, and the graph indeed crosses the horizontal axis.$

Shuuya · May 16, 2016

eyeseeyou said:
And now Integrand comes to the rescue after so long...lol

because the answer sheet shows that it doesn't cross into the negative side

You mean that t is greater than/equal to 0 right? Because time can't be negative.

InteGrand · May 16, 2016

Shuuya said:
You mean that t is greater than/equal to 0 right? Because time can't be negative.

Ah. I'd thought eyeseeyou was referring to crossing the horizontal axis. Yeah, should only draw it for t ≥ 0.

eyeseeyou · May 16, 2016

Shuuya said:
You mean that t is greater than/equal to 0 right? Because time can't be negative.

yeah...I think so

eyeseeyou · May 16, 2016

InteGrand said:
Ah. I'd thought eyeseeyou was referring to crossing the horizontal axis. Yeah, should only draw it for t ≥ 0.

Oh yeah wasn't aware of that lol

eyeseeyou · May 16, 2016

Did no one help me with my last question!!!

InteGrand · May 16, 2016

eyeseeyou said:
Did no one help me with my last question!!!

Looks like no-one did yet. But that's a 4U type question (I think). Are you already up to 4U?

eyeseeyou · May 16, 2016

InteGrand said:
Looks like no-one did yet. But that's a 4U type question (I think). Are you already up to 4U?

No I'm not Integrand

eyeseeyou · May 16, 2016

Did anyone help me with Q2 or Q3?

eyeseeyou · May 16, 2016

BTW thanks for the help guys

Really appreciated it

If I didn't ask I wold've been dead

InteGrand · May 16, 2016

eyeseeyou said:
BTW thanks for the help guys

Really appreciated it

If I didn't ask I wold've been dead

Why would you be dead if you hadn't asked? You could've always asked your teacher or someone tomorrow if no-one had been able to help here, couldn't you?

davidgoes4wce · May 16, 2016

$Q 4 (a) \frac{dx}{dt}=\frac{1}{1+t}$

$dx= \frac{1}{1+t} \ dt$

$x=ln (1+t) + C$

$When t=0, x=0$

$0=ln 1 + C$

$\therefore C=0$

$x=ln \ (1+t)$

$(b) \frac{dx}{dt}=\frac{1}{1+t} . $ As time (t) $ >0, v>0 $ for all values of 't' $$

$a=\frac{d^{2}x}{dt^{2}}=\frac{-1}{(1+t)^2}$

$Q4 (c) $ Let a=0 $ , 0=\frac{-1}{(1+t)^{2}}$

$$ No values of t are valid, where acceleration is equal to 0.$

leehuan · May 16, 2016

davidgoes4wce said:
$Q4 (c) $ Let a=0 $ , 0=\frac{-1}{(1+t)^{2}}$

$$ No values of t are valid, where acceleration is equal to 0.$

I already did Q4, and it said to find the acceleration when t=0

davidgoes4wce · May 16, 2016

I read the question the other way around. My bad.

Prelim 2016 Maths Help Thread (3 Viewers)

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