# 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 :

pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
what are scalars
show that 1w= 10^7ergs^-1
what's lamin's theorems and it's mathematics representative
if the wavelength is double,what is the frequency of the wave
What are the system of units
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
58asagravitasnal firce
Amar
water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin