# Uniform acceleration vector problem

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 gonna 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? im not so confident

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

my apologies about the thread being posted three times, i dont know why but everytime i edited the thread it just posted a new 1, sorry.

thanks for the reply BerryBoy, ill apply that formula, dont worry bout the graph, thanks for the offer though :)