Question

If acc" of Particle is given by eq" $\vec{a}$ = (2x+1) m/s², where x Jis in mater (m). If at x=1m, v=2m/sec. Then find v as f(x) (Junction of position)

$\vec{a}$ = $\int 2x + 1$

Answer 1

v = $\sqrt{2x^2 + 2x}$

Answer 2

The acceleration of the particle is given by $\vec{a} = (2x+1) \, \text{m/s}^2$. Since the acceleration is given as a function of position $x$, we use the relation $a = v \frac{dv}{dx}$, where $v$ is the velocity of the particle.

So, we have: $v \frac{dv}{dx} = 2x+1$

This is a separable differential equation. We can separate the variables $v$ and $x$: $v \, dv = (2x+1) \, dx$

Now, we integrate both sides. We are given that at $x=1 \, \text{m}$, the velocity $v=2 \, \text{m/s}$. We can use these as the limits of integration. Let the velocity at position $x$ be $v(x)$. $\int_{2}^{v} v' \, dv' = \int_{1}^{x} (2x'+1) \, dx'$

Integrating the left side: $\int_{2}^{v} v' \, dv' = \left[ \frac{(v')^2}{2} \right]_{2}^{v} = \frac{v^2}{2} - \frac{2^2}{2} = \frac{v^2}{2} - 2$.

Integrating the right side: $\int_{1}^{x} (2x'+1) \, dx' = \left[ \frac{2(x')^2}{2} + x' \right]_{1}^{x} = \left[ (x')^2 + x' \right]_{1}^{x}$ $= (x^2 + x) - (1^2 + 1) = x^2 + x - 2$.

Equating the results from both sides: $\frac{v^2}{2} - 2 = x^2 + x - 2$

Add 2 to both sides: $\frac{v^2}{2} = x^2 + x$

Multiply by 2: $v^2 = 2(x^2 + x)$ $v^2 = 2x^2 + 2x$

Taking the square root to find $v$: $v = \pm \sqrt{2x^2 + 2x}$

We are given the initial condition that at $x=1$, $v=2$. Let's check this condition with the derived equation: At $x=1$, $v^2 = 2(1^2 + 1) = 2(1+1) = 2(2) = 4$. So, $v = \pm \sqrt{4} = \pm 2$. Since the given initial velocity is $v=2$ (positive), we choose the positive root. $v(x) = \sqrt{2x^2 + 2x}$.

This function represents the velocity as a function of position $x$ for the particle, consistent with the given acceleration and initial condition. The domain of $x$ for which the velocity is real is where $2x^2+2x \ge 0$, which is $2x(x+1) \ge 0$, meaning $x \le -1$ or $x \ge 0$. The initial condition is at $x=1$, which is in the region $x \ge 0$. Since the initial velocity is positive and the acceleration $2x+1$ is positive for $x > -1/2$, the velocity will remain positive as the particle moves in the positive direction from $x=1$.