Question

The velocity of a body at time t=$0$ is in the north-east direction and it is moving with and acceleration of $2m/s$ directed towards the south, The magnitude and direction of the velocity of the body after $5$ sec will be.

Answer 1

The speed of the particle moving towards the east (let’s say x-direction) will remain unchanged… as the direction of the force is perpendicular to the direction of the force applied.

$V_x=2m/sec$Vx=m/sec

The uniform acceleration of the particle toward the north (let’s say y-direction) will be :

$a=\frac{3}{4}m/sec$ $F=a\times mass$a=3/2m/sec2[F=a×mass]

So, the velocity of the particle in the y-direction at time t will be :

$V_y=a\times t=3\times \frac{t}{2}m/sec$Vy=a×t=3×t/m/sec

So, at time t after the application of the force, the velocity of the particle will be:

$V=V_{2}y+V_{2}x...............m/sec$V=Vy2+Vxm/sec

The displacement of the particle at a time interval of dt will be:

$ds=V\times dt$ds=V×dt

So, total displacement of the particle after the application of the force is:

$S=\int 20V\times dt$

$=(\int 204+\frac{9t_2}{4})dt$S=∫02V×dt

=∫024+(9t2/4)×dt

so, displacement will be $5.2meter$.