Question

If velocity of particle moving along x axis is given by eq$^n$ v=3x$^2$, where x is position in m and v is velocity in m/sec. If at t=1 sec. x=1m. Then find position at t=2sec.

Answer 1

-0.5

Answer 2

The velocity of the particle moving along the x-axis is given by the equation $v = 3x^2$.

We know that velocity is the rate of change of position with respect to time, i.e., $v = \frac{dx}{dt}$. Substituting the given expression for $v$, we get: $\frac{dx}{dt} = 3x^2$

This is a first-order separable differential equation. We can separate the variables $x$ and $t$: $\frac{dx}{x^2} = 3 dt$

Now, integrate both sides of the equation. $\int \frac{dx}{x^2} = \int 3 dt$ The integral of $x^{-2}$ with respect to $x$ is $-x^{-1}$, and the integral of a constant $k$ with respect to $t$ is $kt$. $-\frac{1}{x} = 3t + C$ where $C$ is the constant of integration.

We are given that at $t=1$ sec, the position is $x=1$ m. We can use this initial condition to find the value of $C$. Substitute $t=1$ and $x=1$ into the equation: $-\frac{1}{1} = 3(1) + C$ $-1 = 3 + C$ $C = -1 - 3 = -4$

So, the equation relating position $x$ and time $t$ is: $-\frac{1}{x} = 3t - 4$

We need to find the position of the particle at $t=2$ sec. Substitute $t=2$ into the equation: $-\frac{1}{x} = 3(2) - 4$ $-\frac{1}{x} = 6 - 4$ $-\frac{1}{x} = 2$

Now, solve for $x$: $\frac{1}{x} = -2$ $x = -\frac{1}{2}$

So, the position of the particle at $t=2$ sec is $-0.5$ m.