Q. 21

# A particle starts

Answer :

Here the velocity and acceleration of particle are given in vector form and are unit vectors along X axis and Y axis respectively

i.e. they are vectors with magnitude unity and are representing X direction, and Y direction, each vector in x-y plane can be resolved into two components one along with x-direction and other along the y-direction, the vector quantity is the linear combination of both the components

the initial velocity of the particle is

i.e its magnitude is 10.0 m/s in y direction

since coefficient of is zero so magnitude in x direction is zero

the acceleration of the particle is

i.e. the magnitude in x-direction is 8.0 m/s and magnitude along y-direction is 2.0 m/s

As we know

Where,
u is the velocity vector at T=0 s
v is the velocity factor of the particle at time T

(a) since the particle started from the origin, so its displacement in x-direction will give its new x coordinate and displacement in y-direction will give its new y coordinate, so the displacement of the particle in the x-direction is

Sx = 16m

From the above equation.

comparing the x-component with Sx, we get

i.e. when x co-ordinate of a particle is 16m when time t = 2s
Now putting the value T=2 sec

y=10×2 +22 =24 m

(b) Speed is the magnitude of the velocity of the particle so we will first find the component of velocity in x and y-direction using the equation of motion.
Now using the final velocity equation, we get

The speed of the particle

So the speed of the particle is 21.26 m/s

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