First slide

Graphical representation-Motion in 1 d

Question

Particle moving in a straight line is shown in fig. Corresponding acceleration versus velocity graph will be

Difficult

A
B
C
D

Solution

$acceleration a = \frac{d v}{d t}$
$= (\frac{d v}{d s}) (\frac{d s}{d t})$
$= (\frac{d v}{d s}) . V \dots \dots \dots . . (1) (∵ \frac{d s}{d t} = V)$
From given graph
$\frac{d v}{d s} = \frac{10}{10} = 1 S^{- 1}$
$∴$ From Equation ……..(1)
$\begin{array}{l} a = 1 \times V \\ \Rightarrow a = V \end{array}$
Therefore acceleration versus velocity graph will be a straight line passing through origin with slope 1S^-1.

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