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:
Step-by-step explanation:
-1
0.1
.66
1.14
by the numbers it woud be
-1
0.1
2/3
8/7
Answer:
C. It would be shifted up.
Step-by-step explanation:
We base it off the y-intercept. Since -7 is <em>less</em><em> </em><em>than</em><em> </em>1, and our new function has a positive y-intercept, it would indeed be shifted up.
I am joyous to assist you anytime.
Answer:
I WAS BORN TO FLEX DIMONDS ON MY NECK I LIKE MORNING CHECKS I LIKE MORING S**
Step-by-step explanation: