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:
The length between the two points is RT = 1.3 units
Step-by-step explanation:
The coordinates of the point R and T are R(2,1.2) and T(2,2.5).
Now, by DISTANCE FORMULA:
The distance between two coordinates P and Q with coordinate P(a,b) and Q(c,d) is given as:
So, here the distance RT =
or, RT = 1.3 units
Hence, the length between the two points is RT = 1.3 units