Question

Help me anyone on physics?

Answer 1

We get `dx=3 \ dt` and `dy=6x \ dt = 2x \ dx` or `y=x^2+C` through the origin, choice (1).

Explanation:

`V=3i+6xj`

Given a point on the curve `X=x i+yj`, a short time `dt` later the small vector `dX` we add to `X` to get a nearby point on the curve is given by

`dX = V dt`

We write this in terms of the components,

`dx = 3 \ dt`

`dt = dx/3`

`dy = 6x \ dt = 6x \ dx / 3= 2 x \ dx`

Integrating,

`y = x^2 + C`

The launch point is the origin, so `(0,0)` is on the curve, so `C=0.`

Choice (1).

Answer 2

(1)

Explanation:

Given `vecv=3hati+6hatj`
As `xand y` are orthogonal to each other these can be treated separately. Splitting velocity we get

`v_x=3`

We know that velocity means `x` changes by `dx` in infinitesimal time `dt`. Hence above equation becomes

`dx=3dt`

Integrating both sides with respective variables we get

`int\ dx=int\ 3dt`
`=>x=3t+c`
where `c` is constant of integration.

To evaluate `c` we use initial condition. Given `x=0` at `t=0`. Inserting this in above equation we get `c=0`. Hence our equation becomes

`x=3t` .........(1)

Similarly for `v_y`

`v_y=6x`

We know that velocity means `y` changes by `dy` in infinitesimal time `dt`. Hence above equation becomes

`dy=6xdt`

Using (1)

`dy=6(3t)dt`
`dy=18tdt`

Integrating both sides with respective variables we get

`int\ dy=18int\ tdt`
`=>y=18xxt^2/2+c_1`
`=>y=9t^2+c_1`
where `c_1` is constant of integration.

To evaluate `c_1` we use initial condition. Given `y=0` at `t=0`. Inserting this in above equation we get `c_1=0`. Hence our equation becomes

`y=6t^2` .........(2)

Eliminating `t` from (1) and (2) we get

`y=x^2`

Answer 3

Alternate solution in this case.
(1)

Explanation:

Given `vecv=3hati+6hatj`
As `xand y` are orthogonal to each other these can be treated separately. Splitting velocity we get

`v_x=3`

Using definition of velocity we get

`dotx=3`

Integrating both sides with respect to time we get

`int\ dotx\ dt=int\ 3\ dt`
`=>x=3t+c`
where `c` is constant of integration.

To evaluate `c` we use initial condition. Given `x=0` at `t=0`. Inserting this in above equation we get `c=0`. Hence our equation becomes

`x=3t` .........(1)

Similarly for `v_y`

`v_y=6x`
`=>dot v=6x`

Using (1)

`doty=6(3t)dt`
`=>doty=18t`

Integrating both sides with respect to time we get

`int\ doty\ dt=18int\ t\ dt`
`=>y=18xxt^2/2+c_1`
`=>y=9t^2+c_1`
where `c_1` is constant of integration.

To evaluate `c_1` we use initial condition. Given `y=0` at `t=0`. Inserting this in above equation we get `c_1=0`. Hence our equation becomes

`y=6t^2` .........(2)

Eliminating `t` from (1) and (2) we get trajectory of particle as

`y=x^2`

.-.-.-.-.-.-.-.-.-
These solutions have been posted keeping in my that

`d/dt` in an expression `dx/dt` is an operator.
`because dx/dt` is actually `d/dt(x)`

Mathematicians feel that it is not good a idea to break the operator into numerator and denominator as we generally resort to in Physics problems.