# Particle moving under influence of a constant force is given by V =√|4-2X| where X is magnitude of displacement of the particle At t=0 initially the particle is noticed to be moving towards east.The distance travelled by the particle in first 5seconds is?

Dec 6, 2017

Total Distance Travelled:
$\setminus \quad {S}_{\text{tot}} = {S}_{1} + {S}_{2} + {S}_{3} = \left(4 + \frac{5}{2}\right) m = \frac{13}{2} \setminus \quad m$

#### Explanation:

Note: The question fails to provide some necessary information. But it looks like it is an oversight error. Initial displacement of the body must be stated. Without that information it is not possible to solve this problem.

Assumptions: Assuming that $v \left(x\right)$ is the speed of the particle for a displacement magnitude of $x$. I am taking the initial displacement ${x}_{0}$ as zero (since it is not given and is necessary)

v(x) = \sqrt{4-2x}; \qquad v_0 = \sqrt{4-2x_0}=2\quad m.s^{-1}

Acceleration: Force is mentioned as constant. So it would mean constant acceleration too.

a = (dv)/(dt) = \frac{\delv}{\delx}.\frac{dx}{dt} = v.\frac{\delv}{\delx};

$a = \setminus \sqrt{4 - 2 x} . \left\{\setminus \frac{- 2}{2 \setminus \sqrt{4 - 2 x}}\right\} = - 1 \setminus \quad m . {s}^{- 2}$

Equation of Motion (constant acceleration): Displacement is given by the equation -

x(t)-x_0 = v_0t + 1/2at^2;
x_0 = 0m; \qquad v_0=2ms^{-1}; \qquad a=-1m.s^{-2}

${x}_{e} \left(t\right) = \left(+ 2 \frac{m}{s}\right) t - \left(\frac{1}{2} \frac{m}{s} ^ 2\right) {t}^{2}$

We have to distinguish between distance travelled and displacement. Because the object has negative acceleration, it may turn around and start moving in the western direction.

First find the turn around time ($T$). At the turn-around point the speed is zero.

v_e(t) = v_0 + at; \qquad v(t) = 2 - t; \qquad v(T) = 0 \qquad \rightarrow T = 2s

The distance travelled in this time is ${S}_{1} = {x}_{e} \left(T\right) = 2 m$

After turning around the body will travel for another $2 s$ to reach the starting point, travelling a total distance of ${S}_{1} + {S}_{2} = 4 m$ in $4 s$.

When it reached the starting point its speed will be $2m/s$ pointing in the western direction. Now it will starts travelling in the western direction. Reset the initial velocity to $- 2 m {s}^{- 1}$ and calculate the distance it travels in the last 1 second.

x_w(t) = -v_0t-1/2at^2 = (-2m/s)t-(1/2 m/s^2)t^2;

${S}_{3} = | {x}_{w} \left(1\right) | = \frac{5}{2} m$

Total Distance Travelled:
$\setminus \quad {S}_{\text{tot}} = {S}_{1} + {S}_{2} + {S}_{3} = \left(4 + \frac{5}{2}\right) m = \frac{13}{2} \setminus \quad m$