Question

11. A particle starts from origin at t=0 with a velocity of 10-0 jms-1 and moves in the x-y plane with a constant
acceleration of (8.0 î + 2-0 j) ms-2
(a) At what time is the x-coordinate of the particle 16 m ? What is the y-coordinate of the particle at that
time?​

Answer 1

Answer:

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^2

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

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^2

here, u_y=10m/s^2, a_y=2m/s^2 , 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_xt

= 0 + 8i × 2 = 16i m/s

velocity of particle in y direction after 2 sec ,

v_y=u_y+a_yt

= 10 j + 2j × 2= 14j m/s

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

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

= 21.26m/s

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Answer 2

Answer:

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