Question

The figure shows the velocity and acceleration of a particle at the initial moment t=0 of its motion the acceleration vector of the particle remain constant.
Find
1.The instant of time after t=0 at which the speed of the particle is 8 m/s again
2.the instant of time at which the acceleration and velocity vector of th particle are mutually perpendicular
3.the magnitude of displacement of the particle in time interval t=0 to t=4

Answer 1

1. The speed will be 8 m/s after 4 seconds.

2. Acceleration and velocity vector of the particle are mutually perpendicular after 2 seconds

3. the magnitude of displacement of the particle in time interval t=0 to t=4 is 16√3 m

Explanation:

The figure is attached

At t=0, the velocity vector can be written as

$\vec u=8\hat i$

At t=0, the acceleration vector can be written as

$\vec a =-2\cos 60^\circ\hat i-2\sin 60^\circ\hat j$ ....... (1)

$\implies \vec a=-\hat i-\sqrt{3}\hat j$  ................. (2)

(1) Let after time t the speed is again 8 m/s

The velocity in x direction after time t

By first equation of motion

$v_x=u_x+a_xt$

$\implies v_x=8-t$

Similarly, velocity in y direction after time t

$v_y=u_y+a_yt$

$\implies v_y=0-\sqrt{3}t$

Therefore, the speed will be v = 8

$v=\sqrt{v_x^2+v_y^2}$

$\implies 8 = \sqrt{(8-t)^2+(-\sqrt{3})^2}$

$\implies 64=64-16t+t^2+3t^2$

$\implies 4t^2-16t=0$

$\implies 4t(t-4)=0$

$\implies t=0,4$

Therefore, after time t=0, the next the speed will be 8 m/s after 4 seconds.

(2)The acceleration is constant therefore, its vector will be what is computed in eq (2)

The velocity vector after time t is given by

$\vec v=v_x\hat i+v_y\hat j$

$\implies \vec v=(8-t)\hat i-\sqrt{3}t\hat j$

If acceleration vector and velocity vectors are perpendicular

Then

$\vec a.\vec v=0$

$\implies (-\hat i-\sqrt{3}\hat j).((8-t)\hat i-\sqrt{3}t\hat j)=0$

$\implies -8+t+3t=0$

$\implies 4t-8=0$

$\implies t=2$ seconds

Therefore, after 2 seconds, acceleration and velocity vector of th particle are mutually perpendicular

(3) From t=0 to t=4, displacement in x direction

$s_x=u_xt+\frac{1}{2}a_xt^2$

$\implies s_x=8\times 4+\frac{1}{2}\times (-1)\times 4\times 4$

$\implies s_x=32-8=24$ m

Similarly, displacement in y direction

$s_y=u_yt+\frac{1}{2}a_yt^2$

$\implies s_y=0+\frac{1}{2}(-\sqrt{3})\times 4\times 4$

$\implies s_y=-8\sqrt{3}$

Thus, the magnitude of displacement

$s=\sqrt{s_x^2+s_y^2}$

$\implies s=\sqrt{24^2+(-8\sqrt{3})^2}$

$\implies s=\sqrt{24\times 24+8\times 8\times 3}$

$\implies s=8\sqrt{9+3}$

$\implies s=8\sqrt{12}$

$\implies s=16\sqrt{3}$

Therefore, the magnitude of displacement of the particle in time interval t=0 to t=4 is 16√3 m

Hope this answer is helpful.

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