A particle starts from origin and moves with velocity (81)ms-1 at time t = 0. The particle moves in x-y plane with an acceleration of a = (-2i+ 4ſ)ms? The position co-ordinates at which velocity of the particle becomes parallel to y-axis, is ?

Answers 1

Explanation:

initial velocity of particle, u = 10j m/s

acceleration of particle , a = (8i + 2j) m/s²

displacement of particle in x direction , x = 16m

(a) so, use formula x=u_xt+\frac{1}{2}a_xt^2x=u

here, u_x=0,a_x=8m/s^2u

=0,a

=8m/s

so, 16 = 0 + 1/2 × 8 × t²

or, 16 = 4t² => t = 2

hence, after 2 sec particle moves 16m in x direction.

now we have to find y - co-ordinate of particle at 2 sec.

so, use formula, y=u_yt+\frac{1}{2}a_yt^2y=u

here, u_y=10m/s^2u

=10m/s

, a_y=2m/s^2a

=2m/s

, t = 2sec

so, y = 10 × 2 + 1/2 × 2 × 2²

y = 20 + 4 = 24m

(b) velocity of particle in x direction after 2 sec, v_x=u_x+a_xtv

= 0 + 8i × 2 = 16i m/s

velocity of particle in y direction after 2 sec ,

v_y=u_y+a_ytv

= 10 j + 2j × 2= 14j m/s

hence, speed , |v| = \sqrt{v_x^2+v_y^2}

= \sqrt{16^2+14^2}

+14

= 21.26m/s