Question

e. A particle starts moving from origin such that its x and y
components of velocity are given as V = 5y and V =4. The
particle will follow-
(1) Straight
(2) Parabola
(3) Circle
(4) Ellipse


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Answer 1

Answer:

The velocity of particle is given :-

V=(8t-2)i+2j

This means that the particle is moving with a constant velocity 2 in y-direction , hence it will have no acceleration in this direction.

In x-axis , the velocity of the particle is a function of time , i.e. , the velocity of particle in x direction is changing with time , hence it will have an acceleration in this direction.

NOW :- Distance travelled by particle in y-axis in time t is equal to 2t.

Therefore , we can write :-

y=2t…..(1)

NOW , for finding the distance travelled in

X-direction , we have to integrate velocity in x-axis with respect to time , which is done as follows :-

SEE TO IMAGE

So, the displacement in x-direction comes out to be a function of time.

x=4t^2–2t+c …(2)

NOW, it is given that , at time t=2 seconds , the particle is at (14,4)

Putting , time t=2 seconds , in equation….(2) , we get :-

14=4*(2^2)-2(2)+c

OR, 14=4*4–4+c

OR, 14=16–4+c

OR, 14=12+c

OR, 14–12=c

Therefore, the value of constant 'c' comes out to be 2.

Therefore ,

X=4t^2–2t+2…(3)

From equation…(1) , we get:-

t=y/2

Putting value of t in equation…(3) we get:-

x=4*(y/2)^2–2(y/2)+2

OR, x=4*(y^2/4)-y+2

OR, x=y^2-y+2

Hence, the equation of the path of the particle is

x=y^2-y+2.

I hope this answer might help you.

THANK YOU.

Explanation:

Answer 2

Answer:

Answer of this question is parabola...