app of callculus URGENT (1 Viewer)

hollaholla · Aug 17, 2009

Really been on this for a longggg timee.....

A particle is accelerating according to the equation A= (3t +1)^2 ms.
If the particle is initially at rest 2m to the left of 0, find its displacement after 4seconds.

Zak Ambrose · Aug 17, 2009

A = 9t^2 + 6t + 1
V = 3t^3 + 3t^2 + C
when t = 0, V=0
V = 3t^3 + 3t^2
D = [(3t^4)/4] + t^3 + C
when t = 0, V = 2
D = [(3t^4)/4] + t^3 +2
when t = 4
D = [(12^4)/4] + 4^3 + 2

Drongoski · Aug 17, 2009

acceleration is given in ms^-2 and calculus is spelt as shown.

I'll now make an attempt on the question.

$\100dpi a = \frac{\mathrm{d^{2}}x }{\mathrm{d} t^{2}}\ =\ (3t+1)^{2}\ \ ms^{-2}\\ \\ \therefore \ \ v\ =\ \frac{\mathrm{d} x}{\mathrm{d} t}\ \ =\ \ \int (3t+1)^{2}\ dt\ \ =\ \ \frac{1}{9}(3t+1)^{3}\ +\ C\\ \\ @\ t=0, v=0\ \Rightarrow \ \frac{1}{9}(3\times 0+1)^{3}+C\ =\ 0\ \Rightarrow \ C=-\frac{1}{9}\\ \\ \therefore \ \ v\ =\ \frac{1}{9}(3t+1)^{3}\ -\ \frac{1}{9}\\ \\ \therefore \ \ x\ \ =\ \ \int \frac{\mathrm{d} x}{\mathrm{d} t}\ dt \ \ =\ \ \int \left [\frac{1}{9} (3t+1)^{3}-\frac{1}{9} \right ]\ dt\\ \\ =\ \ \frac{1}{108}(3t+1)^{4}-\frac{t}{9}\ +\ D\\ \\ @\ t=0, \ x=-2\ \Rightarrow \ \ \frac{1}{108}(0+1)^{4}-\frac{0}{9}+D\ =\ -2\ \Rightarrow \ D\ =\ -(2+\frac{1}{108})\\ \\ \therefore \ \ x\ =\ \frac{1}{108}(3t+1)^{4}-\frac{t}{9}-(2+\frac{1}{108})\\ \\ @\ t=4:\ \ x\ =\ \ \frac{1}{108}(3\times 4+1)^{4}-\frac{4}{9}-(2\frac{1}{108})\ \ m\\ \\ You\ work\ this\ out.$

Edit

Oh how silly of me; should have just expanded acceleration like Zak - life would have been much easier.

hollaholla · Aug 17, 2009

Drongoski said:
acceleration is given in ms^-2 and calculus is spelt as shown.

I'll now make an attempt on the question.

$\100dpi a = \frac{\mathrm{d^{2}}x }{\mathrm{d} t^{2}}\ =\ (3t+1)^{2}\ \ ms^{-2}\\ \\ \therefore \ \ v\ =\ \frac{\mathrm{d} x}{\mathrm{d} t}\ \ =\ \ \int (3t+1)^{2}\ dt\ \ =\ \ \frac{1}{9}(3t+1)^{3}\ +\ C\\ \\ @\ t=0, v=0\ \Rightarrow \ \frac{1}{9}(3\times 0+1)^{3}+C\ =\ 0\ \Rightarrow \ C=-\frac{1}{9}\\ \\ \therefore \ \ v\ =\ \frac{1}{9}(3t+1)^{3}\ -\ \frac{1}{9}\\ \\ \therefore \ \ x\ \ =\ \ \int \frac{\mathrm{d} x}{\mathrm{d} t}\ dt \ \ =\ \ \int \left [\frac{1}{9} (3t+1)^{3}-\frac{1}{9} \right ]\ dt\\ \\ =\ \ \frac{1}{108}(3t+1)^{4}-\frac{t}{9}\ +\ D\\ \\ @\ t=0, \ x=-2\ \Rightarrow \ \ \frac{1}{108}(0+1)^{4}-\frac{0}{9}+D\ =\ -2\ \Rightarrow \ D\ =\ -(2+\frac{1}{108})\\ \\ \therefore \ \ x\ =\ \frac{1}{108}(3t+1)^{4}-\frac{t}{9}-(2+\frac{1}{108})\\ \\ @\ t=4:\ \ x\ =\ \ \frac{1}{108}(3\times 4+1)^{4}-\frac{4}{9}-(2\frac{1}{108})\ \ m\\ \\ You\ work\ this\ out.$

Edit

Oh how silly of me; should have just expanded acceleration like Zak - life would have been much easier.

lol gosh totally forgot about expansion...thanks guys...!.

Aquawhite · Aug 17, 2009

rofl... I would never consider not expanding that. Makes life a hell lot easier to simply expand a binomial.

app of callculus URGENT (1 Viewer)

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