The velocity-displacement graph of a particle moving along a straight line is shown [2005]

The most suitable acceleration-displacement graph will be

Question

The velocity-displacement graph of a particle moving along a straight line is shown [2005]

The most suitable acceleration-displacement graph will be

Smart Achievers · Accepted Answer

(1)

$The equation for the given v - x graph is$

$v = - \frac{v_{0}}{x_{0}} x + v_{0} ...(i)$

$\frac{d v}{d x} = - \frac{v_{0}}{x_{0}}$

$∴$ $a = v \frac{d v}{d x} = - \frac{v_{0}}{x_{0}} \times v = - \frac{v_{0}}{x_{0}} [- \frac{v_{0}}{x_{0}} x + v_{0}] from (i)$

$\Rightarrow a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}} ...(ii)$

$On comparing equation (ii) with equation of a straight line$

$y = m x + c$

$we get m = \frac{v_{0}^{2}}{x_{0}^{2}} = +ve,$

$i.e., \tan θ = + ve, i.e., θ is acute.$

$Also c = - \frac{v_{0}^{2}}{x_{0}^{2}}$ ,

$i.e., the y-intercept is negative. Hence graph (a) correctly depicts corresponding a - x graph.$

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