Recent content by LockeZz

I shall check back from the beginning the see if any mistake again.. thanks for the guidance by the way.. i really appreciate on your help.. Thank you very much :)

hm.. does that mean that the negative sign represent a mistake in the expression?

i think i can convert to tangent: x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}}))) then simplify again: x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))

after the substitution, i get : x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))}) since t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}...

get rid of the x0 ad substitute the t1?

ok.. the logarithm still can be simplify : x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)}) so.. for t0=0.. i...

i have check on the table.. is cosine.. x-x_0=-\frac{m}{\beta}(\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))

hm.. i think i know where did i went wrong, i miss out the secant: x-x_0=-\frac{m}{\beta}(\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))

By integration with substitution in which i assign the limit for dx(x,x0)and limt for dt(t1,t0) : x-x_0=-\frac{m}{\beta}(\ln(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)-\ln(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0))

i give it a try.. and i get this : v=\sqrt{\frac{\alpha}{\beta}}tan(\arctan(v_0\ sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t)

im sorry.. i don't quite understand what u mean.. can explain in another word?

i was stucked at here... how to invert the subtraction of two arctangent? \frac{\sqrt{\alpha\beta}}{m}t=\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\arctan(v\sqrt{\frac{\beta}{\alpha}})

i had try to substitute the v0 as infinity and then invert the tangent to get v(t) which i gained : v=\sqrt{\frac{\alpha}{\beta}}(tan(\frac{\pi}{2}-\frac{t}{m}(\sqrt{\alpha\beta}))) then i integrate it by using the substitution method ( taking x=distance traveled and x0 as starting point...

oops another careless mistake... so the time for maximum deceleration is this: t=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}} Do i need to find the acceleration function?

alright.. i have substitute the final velocity to be 0 and initial velocity which is very large within the arctangent will result in 900 which is half of the pi. so the time for maximum deceleration in which v0 approaching 0 will be: t=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}