Author Topic: Vector Calculus help. (Read 753 times) Tweet Share

cheeky_monkey · « **on:** September 15, 2010, 08:56:35 pm »

Could someone please work out this question.
V0=20i-10j+40k
alltitude = 100m
r0=100k and acceleration=-10k

a) v(t)
b) r(t)
c) time and height above ground when velocity and acceleration are at right angles.

m@tty · « **Reply #1 on:** September 15, 2010, 09:20:38 pm »

$a=-10\bold{k}$

$v(t)=\int a dt$

$v(t)=-10t\bold{k}+V_0$

$v(t)=20\bold{i}-10\bold{j}+(40-10t)\bold{k}$

$r(t)=\int v(t) dt$

$r(t)=20t\bold{i}-10t\bold{j}+(40t-5t^2)\bold{k}+r_0$

$r(t)=20t\bold{i}-10t\bold{j}+(100+40t-5t^2)\bold{k}$

c)

$a\cdot v=0$

$-10(40-10t)=0 \implies t=\frac{40}{10}=4$

$r(4)=80\bold{i}-40\bold{j}+180\bold{k}$

I'm assuming 'height above the ground' is given by the value in the k direction.

Therefore height = 180m.

sajib_mostofa · « **Reply #2 on:** September 15, 2010, 09:22:32 pm »

a) $\int -10k dt = C_1i + C_2j + (-10tk + C_3)k<br />$
Sub $V(0) = 20i - 10j + 40k$ and solve:

So $C_1 = 20, C_2 = -10, C_3 = 40<br />$
\therefore $V(t) = 20i -10j + (40-10t)k<br />$

cheeky_monkey · « **Reply #3 on:** September 15, 2010, 09:23:43 pm »

Thanks so much

sajib_mostofa · « **Reply #4 on:** September 15, 2010, 09:23:47 pm »

Damn how do you do it so quickly Matty!

m@tty · « **Reply #5 on:** September 15, 2010, 09:25:42 pm »

Haha, I am used to LaTeX from procrastinating on here all year, rather than doing work.

Search the forums now!

We have moved!

Author Topic: Vector Calculus help. (Read 753 times) Tweet Share

cheeky_monkey

Vector Calculus help.

m@tty

Re: Vector Calculus help.

sajib_mostofa

Re: Vector Calculus help.

cheeky_monkey

Re: Vector Calculus help.

sajib_mostofa

Re: Vector Calculus help.

m@tty

Re: Vector Calculus help.

Recent Posts

User:
Password: