2.50 x 4 = 10
20 - 10 = 10
so his change would be $10
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 .
-6p^3(3p^2 +5p -1) = -18p^5 -30p^4 +6p^3
so this is the right answer sure
Question: Where is the y2? Please be more specific!
Since the denominators are the same, just subtract the numerators.
The answer is 3/10.