Given :
A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v=2i-4tj .
To Find :
A. The vector position of the particle at any time t .
B. The acceleration of the particle at any time t .
Solution :
A )
Position of vector v is given by :

B )
Acceleration a is given by :

Hence , this is the required solution .
Answer:
64.4
Step-by-step explanation:
In one of the trig formulas, it states sin <a/A = sin<b/B = sin<c/C
So we have in this case:
sin(70)/12 = sin x/14 Note: x is just a variable for the angle <ABC
14(sin70)/12 = sin x
sin^-1(14(sin70)/12) = x
x=64.4
Your welcome, and comment if you have any questions! :D
It’s 90 degrees because it’s a right angle