Name: Rectilinear Motion and Its Example
Uploaded: 2019-07-19
Duration: 11 min 6 s
Description: What is the similarity between the motion of people marching and the motion of a ball rolling down an inclined plane? Let's get an answer in this video.

Question

Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.

Accepted Answer

Consider a particle moving in a straight line with an initial velocity u and constant acceleration a. The v - t graph is a straight line with slope a and y-intercept u. With u and a both positive, the graph is as shown in Fig. $3.9$ .

The slope of the line gives the acceleration of the particle,

$A = \frac{∆ v}{t} = \frac{v - u}{t - 0} = \frac{v - u}{t}$

$∴ v = u + at$

This is the first equation of motion ( or the first kinematical equation).

The area under the straight line for a time

Interval $∆ t = t - 0$ , as shown by the shaded region,

area of rect. $OACD + area of ∆ ABC$

$= (OA \times AC) + \frac{1}{2} (AC \times BC)$

$= ut + \frac{1}{2} t (v - u)$

$= ut + \frac{1}{2} {at}^{2} (∵ v - u = at)$

$= s$ , distance covered in the elapsed time t.

^{$∴ s = ut + \frac{1}{2} {at}^{2}$}

This is the second equation of motion ( or the second kinematical equation).

[Alternatively, the shaded area $=$ area of the trapezium

$OABD = \frac{1}{2} (OA + BD) \times OD$

$= \frac{1}{2} (u + v) t$

$= v_{av} t$ $\{∵ v_{av} = \frac{1}{2} (u + v)\}$

= distance covered in the elapsed time t ]

$∴ s = \frac{1}{2} (u + v) (\frac{v - u}{a})$

substituting for t from the first kinematical equation.

$∴ s = \frac{v^{2} - u^{2}}{2 a}$

∴ v² = u² + 2as $∴ v^{2} = u^{2} + 2 as$

This is the third equation of motion ( or the third kinematical equation).

Derive equations of motion graphically for a particle having uniform acceleration, moving along a straight line.

Important Questions on Motion in a Plane

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