# 2.7 Graphs of motion with constant acceleration  (Page 5/6)

 Page 5 / 6

Case 4: Initial position and origin of reference are same. Initial velocity is positive i.e. it is directed in reference direction. The velocity and acceleration are in same direction.

The particle is accelerated so long force causing acceleration is applied on the particle.

The segement OF is typical graph of free fall of particle under gravity, considering OF as the height of fall and downward direction as positive direction. Only differing aspect is that particle has initial velocity. Neverthless, the nature of curve of free fall is similar. Note that speed of the particle keeps increasing till it hits the ground.

Given ${x}_{0}=0\phantom{\rule{1em}{0ex}}m;\phantom{\rule{1em}{0ex}}u=15\phantom{\rule{1em}{0ex}}m/s;\phantom{\rule{1em}{0ex}}a=10\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Find the time instants when the particle is at origin of reference. Also find the time when velocity is zero.

The position of the particle with respect to origin of reference is given by :

$x=ut+\frac{1}{2}a{t}^{2}$

In order to determine time instants when the particle is at origin of reference, we put x=0,

$0=15t+\frac{1}{2}X10{t}^{2}$ $⇒{t}^{2}+3t=0$ $⇒t=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}-3\phantom{\rule{1em}{0ex}}s$

Neglecting negative value of time,

$⇒t=0$

The particle is at the origin of reference only at the start of motion. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

$⇒0=15+10t$ $⇒t=-1.5\phantom{\rule{1em}{0ex}}s$

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.

## Acceleration is negative (opposite to the reference direction)

In this case, the graph of quadratic equation is a parabola opening downwards as coefficient of squared term ${t}^{2}$ is negative i.e. a>0. The maximum value of expression i.e. x is :

${x}_{\mathrm{max}}=-\frac{D}{4A}=-\frac{{u}^{2}-2a{x}_{0}}{4X\frac{a}{2}}=-\frac{{u}^{2}-2a{x}_{0}}{2a}$

The analysis for this case is similar to the first case. We shall, therefore, not describe this case here.

Given ${x}_{0}=10\phantom{\rule{1em}{0ex}}m;\phantom{\rule{1em}{0ex}}u=15\phantom{\rule{1em}{0ex}}m/s;\phantom{\rule{1em}{0ex}}a=-10\phantom{\rule{1em}{0ex}}m/{s}^{2}$

Find the time instants when the particle is at origin of reference and initial position. Also find the time when velocity is zero.

The position of the particle with respect to origin of reference is given by :

$x={x}_{0}+ut+\frac{1}{2}a{t}^{2}$

In order to determine time instants when the particle is at origin of reference, we put x=0,

$⇒0=10+15t+\frac{1}{2}X-10X{t}^{2}$ $⇒-{t}^{2}+3t+2=0$

$⇒t=\frac{-3±\sqrt{\left\{9-\left(4X-1X2\right)\right\}}}{2X-1}$

$⇒t=\frac{-3±4.12}{-2}=-0.56\phantom{\rule{1em}{0ex}}s\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}3.56\phantom{\rule{1em}{0ex}}s$

We neglect negative value of time. The particle reaches origin of reference only once at t = 2 s. In order to determine time instant when particle is at initial position, we put x=10,

$⇒10=10+15t+\frac{1}{2}X-10{t}^{2}$ $⇒-{t}^{2}+3t=0$ $⇒t=0,\phantom{\rule{1em}{0ex}}3\phantom{\rule{1em}{0ex}}s$

The particle is at the initial position twice, including start point. Now, at the point of reversal of direction, speed of the particle is zero. Putting v = 0 and using v=u+at,

$⇒0=15-10t$ $⇒t=1.5\phantom{\rule{1em}{0ex}}s$

We neglect negative value and deduce that particle never ceases to move and as such there is no reversal of motion.

## Example

Problem : A particle’s velocity in “m/s” is given by a function in time “t” as :

$\begin{array}{l}v=40-10t\end{array}$

If the particle is at x = 0 at t = 0, find the time (s) when the particle is 60 m away from the initial position.

Solution : The velocity is given as a function in time “t”. Thus, we can know its acceleration by differentiating with respect to time :

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