Motion in a plane NCERT Solutions Class 11 Physics - Solved Exercise Question 4.20

Question 4.20:

The position of a particle is given by
r = 3.0t i − 2.0t² j + 4.0 k 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 ?

Solution:

v(t) = (3.0 i - 4.0t j)
a = -4.0 j
The position of the particle is given by:
r = 3.0t i -2.0t² j +4.0 k

Velocity v of the particle is given as:
v = dr / dt = d (3.0t i -2.0t² j + 4.0 k) / dt
∴ v = 3.0 i -4.0t j

Acceleration a of the particle is given by:
v = (3² + (-8)²)^1/2 = (73)^1/2 = 8.54 m/s

Direction, θ = tan^-1(v_y/v_x)
= tan^-1(-8/3)  = -tan^-1(2.667)
= -69.45⁰

The negative sign indicates that the direction of velocity is below the x-axis.