Question

a particle is moving in x y plane. at certain instant of time the components of its velocity and acceleration are as folows - v along x is 3 m/s, v along y is 4 m/s, a along x is 2 m/s^2 and a along y is 1 m/s^2. rate of change of speed at this moment is ? root 10 m/s/2 , 4 m/s^2 , root 5 m/s^2 , 2 m/s^2

Answer 1

For x – direction first,
Applying the eqns. of motion,
v1 = u1 + at
v1 = 8i + 2i*t => (8 + 2t)i ….........................(1)

For y – direction.
acceleration is zero.
thus velocity remains constant.
v2 = u2
v2 = -15j …................(2)

Thus resultant velocity of particle at time t from eqnns. (1) and (2) is given by:

v = v1 + v2
v = (8 + 2t)i – 15j m/s.