Uniform acceleration vector problem

[DB]

i just need some1 to tell me if i did something wrong and if i did a hint in the right direction would be great.

an object accelerates uniformly at 8 m/s^2, 130 degrees for 5 seconds until it reaches a velocity of 20 m/s, 143 degrees.
Calculate: a) the intial velocity b) the displacement at the end of the 5 seconds

a) i decided i was going to use a=\frac{v_2-v_1}{\Delta t}
so i had to turn the vectors to rectangular form, so i worked out their components using trig:

8, 130 degrees = [-5.14, 6.13]
20, 143 degrees = [-16, 12]

so i filled in wat i know:

[-5.14, 6.13] =\frac{[-16, 12] - [x,y]}{5}

5*[-5.14, 6.13]=[-16, 12] - [x,y]

[-25.7, 30.7]-[-16,12]=-[x,y]

\frac{[-9.7, 18.7]}{-1}=[x,y]

[9.7, -18.7]=[x,y]

so know i got the components of the intial velocity, to put them in polar form:

pythagorus
v_1=\sqrt{443.78} \sim 21.1 m/s

using tan inverse i get 27.4 degrees for the angle and because x is negative and y is positive you get 21.1 m/s , West 27.4 degrees North
that wat i get for "a"

b) i chose to use \Delta d= v_2 \Delta t - \frac{1}{2}a \Delta t^2
subsitute:

\Delta d= [-16, 12]*5- \frac{1}{2} [-5.14, 6.13]*25

\Delta d= [-80,60]-[-64.25, 76.63]

\Deta d=[-15.75, -16.63]

so
Delta d= ~ 23 m W 46.5 degrees S

howd i do? I am not so confident

 

[BerryBoy]

I haven't got a calculator with me, but assuming you typed it in right, I agree with you answer to (a). However, your equation for displacement has to use initial velocity (whereas you used final velocity), you can prove this with a velocity-time graph. If you want the graph tell me and I'll add it on in the morning (UK time).

d = V(initial) x T + 0.5 x A x T^2 (sorry about the crude formula)

I think you should be more confident, your thinking is spot on (especially how you approach with questions, just need to learn the formulas) ;-)

Regards,
Sam

 

[DB]

my apologies about the thread being posted three times, i don't know why but everytime i edited the thread it just posted a new 1, sorry.

thanks for the reply BerryBoy, ill apply that formula, don't worry bout the graph, thanks for the offer though :)