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.(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]a=\frac{v_2-v_1}{\Delta t}[/tex]

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:

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

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

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

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

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

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

pythagorus

[tex]v_1=\sqrt{443.78} \sim 21.1 m/s[/tex]

using tan inverse i get 27.4 degrees for the angle and because x is negative and y is positive you get21.1 m/s , West 27.4 degrees North

that wat i get for "a"

b) i chose to use [tex]\Delta d= v_2 \Delta t - \frac{1}{2}a \Delta t^2[/tex]

subsitute:

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

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

[tex]\Deta d=[-15.75, -16.63][/tex]

so

Delta d= ~ 23 m W 46.5 degrees S

howd i do? im not so confident

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# Uniform acceleration vector problem

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