Question

A particle moves in the xy plane starting from time = 0 second and position (1m, 2m) with a velocity of v>=2i-4tj^

. Find
A. The vector position of the particle at any time t,
B. The acceleration of the particle at any time t,​

Answer 1

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 :

`d=\int\limits {v} \, dt\\\\d=\int\limits {(2i-4tj)} \, dt \\\\d=(2t)i+(4t^2)/(2)j\\\\d=(2t)i+(2t^2)j`

B )

Acceleration a is given by :

`a=(dv)/(dt)\\\\a=(2i-4tj)/(dt)\\\\a=(2i)/(dt)-(4tj)/(dt)\\\\a=0-4j\\\\a=-4j`

Hence , this is the required solution .