Answer:
Rule up a graph sheet and text it to me so I can help
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 .
First you need to subtract 750-600= 150.
Then 150 time .35=52.5.
Lastly 52.5+49.99= 102.49
Answer:
It is 4
Step-by-step explanation: