Motion in One Dimension
    6.0 Analysis of motion through graph

6.0 Analysis of motion through graph

The graph helps to analyze and visualize the motion of a body.

Before we start graph, let us understand basic terminologies of the graph.


Slope $\left( {\frac{{dy}}{{dx}}} \right)$

$y=mx+c$ be a straight line.

$m$ is the value of slope

Slope means the inclination of line with respect to the $x$ axis.
$$\tan \theta = m = \frac{{dy}}{{dx}}$$
The first derivative gives the slope.

Note:

1. The slope of a straight line is constant.

2. The slope of a curve is different at a different point.

For finding slope at any point, draw a tangent at that point. The angle made by the tangent with the $x$ axis gives the slope at that point/

Slope at point $A$: ${m_A} = \tan {\theta _1}$

Slope at point $B$: ${m_B} = \tan {\theta _2}$

Slope at point $C$: ${m_C} = \tan {\theta _3}$


The graph helps to analyze and visualize the motion of a body.

Before we start graph, let us understand basic terminologies of the graph.


Slope $\left( {\frac{{dy}}{{dx}}} \right)$

$y=mx+c$ be a straight line.

$m$ is the value of slope

Slope means the inclination of line with respect to the $x$ axis.
$$\tan \theta = m = \frac{{dy}}{{dx}}$$
The first derivative gives the slope.

Note:

1. The slope of a straight line is constant.

2. The slope of a curve is different at a different point.

For finding slope at any point, draw a tangent at that point. The angle made by the tangent with the $x$ axis gives the slope at that point/

Slope at point $A$: ${m_A} = \tan {\theta _1}$

Slope at point $B$: ${m_B} = \tan {\theta _2}$

Slope at point $C$: ${m_C} = \tan {\theta _3}$


Nature of the curve $\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)$

Second derivative tells us the nature of the curve.

ConditionNatureGraphEquationDerivation
$$\frac{{{d^2}y}}{{d{x^2}}} = 0$$Straight line

$$y = mx + c$$$$\frac{{dy}}{{dx}} = m$$$$\frac{{{d^2}y}}{{d{x^2}}} = 0$$
$$\frac{{{d^2}y}}{{d{x^2}}} > 0$$Concave up

$$y = {x^2} + 5$$$$\frac{{dy}}{{dx}} = 2x$$$$\frac{{{d^2}y}}{{d{x^2}}} = 2>0$$
$$\frac{{{d^2}y}}{{d{x^2}}} < 0$$Concave down

$$y = -{x^2} + 5$$$$\frac{{dy}}{{dx}} = -2x$$$$\frac{{{d^2}y}}{{d{x^2}}} = -2<0$$
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