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.Oj) m s².
(a) At what time is the x- coordinate of the particle 16 m? 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, $u_y=10m/s^2$, $a_y=2m/s^2$ and 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:

(a) t = 2, y =24m

(b) 21.26m/s

Explanation:

According to this question

Initial velocity of particle, $u$ = $10j$

Acceleration of particle , $a$ = $(8i + 2j) \frac{m}{s^2}$

Displacement of particle in x direction , x = 16m

(a) So, use formula $x$ = $u_x\ t + \frac{1}{2}\times a_x\times t^2$

Here, $u_x$ = 0 and $a_x$ = $8\frac{m}{s^2}$

So, $16$ = $0 + \frac{1}{2}\times 8\times t^2$

or, 16 = $4t^2$

      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, $y$ =  $u_y\ t + \frac{1}{2}\times a_y\times t^2$

Here, $u_y$ = $10 \frac{m}{s^2}$ , $a_y$ = $2 \frac{m}{s^2}$ and t = 2sec

So, $y$ = $10\times 2 + \frac{1}{2}\times 2\times 2^2$

y = 20 + 4 = 24m

(b) Velocity of particle in x direction after 2 sec,

$v_x$ = $u_x + a_x\times$

= 0 + 8i × 2 = 16i m/s

Velocity of particle in y direction after 2 sec ,

$v_y$ = $u_y + a_y\times t$

= 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