Question

A particle starts from the origin at t = 0 s with a velocity of 10.0 j m/s and moves in the x-y plane with a constant acceleration of (8.0i + 2.0j) m s⁻².a. At what time is the x-coordinate of the particle 16m? What is the y-coordinate of the particle at that time?b. What is the speed of the particle at the time

Answer 1

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

Answer 2

Answer:

(i) T=2 sec

(ii) Speed=21.26 m/s

Explanation:

(i)$u_{x} i + u_{y} j = 10j,$ $[where,u_{x} = 0 ,and ,u_{y} = 10 m/s]$

$a = a_{x} i + a_{y} j = 8i + 2j,$ [where ,$a_{x} = 8 ,and, a_{y} = 2 m/s^{2}$]

$x = u_{x} t + (1/2)a_{x} t^{2}$

$=>16 = 8 t^{2} /2$

t = 2 sec         ... Ans

(ii)$y = u_{y} t + a_{y} t^{2} /2 = 10*2 + 2*2^{2} /2 = 24 m$

$v_{x} = u_{x} + a_{x}t = 16 m/s$

$v_{y} = u_{y} + a_{y} t = 14 m/s$

$Speed = √(v_{x}{^{2} } + v_{y} {^{2}} ) = 21.26 m/s$