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: Walk through writing a general formula for the midpoint between two points. ... I believe you would simply find the differences in x and y from the midpoint to the one ... How would you solve a problem in which you do not know point B but are given ... the line y=x and the curve y=4x-x^2 intersect at the point p and q.
Step-by-step explanation:
I cannot reach a meaningful solution from the given information. To prove that S was always true, you would have to prove that N was always false. To prove that N was always false you would have to prove that L was always false. For the statement (L ^ T) -> K to be true, you only need K to be true, so L can be either true or false.
Therefore, because of the aforementioned knowledge, I do not believe that you can prove S to be true.
4a+8=14-6+4a. 4a+8=8+4a so I think it's B
Yes <span>-(12/-17) is equal to 12/17
I hope this helps.
Have a awesome day. :)</span>
