Question

The position of a particle is given by r = 3.0t ˆi − 2.0t 2 ˆj + 4.0 kˆ m where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find the v and a of the particle? (b) What is the magnitude and direction of velocity of the particle at t = 2.0 s ?

Answer 1

Given: The position vector of a particle is r=3.0t i ^ −2.0 t 2 j ^ +4.0 k ^ m.

a)

The velocity vector is given as,

v= d dt ( r )

By substituting the given values in the above expression, we get

v= d dt ( 3.0t i ^ −2.0 t 2 j ^ +4.0 k ^ ) =3.0 i ^ −4.0t j ^ ms −1

The acceleration vector is given as,

a= d dt ( v )

By substituting the given values in the above expression, we get

a= d dt ( 3.0 i ^ −4.0t j ^ ) =−4.0 j ^ ms −2

b)

The magnitude of velocity of the particle at t=2.0 s is given as,

| v |= v x 2 + v y 2

By substituting the values in the above expression, we get

| v |= ( 3.0 ) 2 + ( −4.0t ) 2 = 9+16×4 =8.54 ms −1

The direction of velocity of the particle at t=2.0 s is given as,

θ= tan −1 ( v y v x )

By substituting the values in the above expression, we get

θ= tan −1 ( −4.0×2 3.0 ) = tan −1 ( −8.0 3.0 ) =−69.4°

Thus, the magnitude of velocity of the particle at t=2.0 s is 8.54  ms -1 and its direction is 69.4° below the x –axis.