Question
The velocity-displacement graph of a particle moving along a straight line is shown
The velocity-displacement graph of a particle moving along a straight line is shown
The most suitable acceleration-displacement graph will be-
A.
B.
C.
D.
Answer :
Solution :
The equation for the given $$v-x$$ graph is
$$\eqalign{ & v = - \frac{{{v_0}}}{{{x_0}}}x + {v_0}.....(i) \cr & \frac{{dv}}{{dx}} = - \frac{{{v_0}}}{{{x_0}}} \cr & \therefore v\frac{{dv}}{{dx}} = - \frac{v}{{{x_0}}} \times v \cr & = - \frac{{{v_0}}}{{{x_0}}}\left[ { - \frac{{{v_0}}}{{{x_0}}}x + {v_0}} \right]{\text{from }}(i) \cr & \therefore a = \frac{{v_0^2}}{{x_0^2}}x - \frac{{v_0^2}}{{{x_0}}}.....(ii)\,\,\left[ {\because a = v\frac{{dv}}{{dx}}} \right] \cr} $$
On comparing the equation (ii) with equation of a straight line
$$\eqalign{ & y = mx + c \cr & {\text{we get }}m = \frac{{v_0^2}}{{x_0^2}} = + ve, \cr & i.e.,\,\tan \theta = + ve,\,\,\,i.e.,\,\,\theta \,\,{\text{is acute}}{\text{.}} \cr & {\text{Also}}\,\,c = - \frac{{v_0^2}}{{x_0^2}}, \cr} $$
$$i.e.,$$ the $$y$$-intercept is negative
The above conditions are satisfied in graph (A).
The equation for the given $$v-x$$ graph is
$$\eqalign{ & v = - \frac{{{v_0}}}{{{x_0}}}x + {v_0}.....(i) \cr & \frac{{dv}}{{dx}} = - \frac{{{v_0}}}{{{x_0}}} \cr & \therefore v\frac{{dv}}{{dx}} = - \frac{v}{{{x_0}}} \times v \cr & = - \frac{{{v_0}}}{{{x_0}}}\left[ { - \frac{{{v_0}}}{{{x_0}}}x + {v_0}} \right]{\text{from }}(i) \cr & \therefore a = \frac{{v_0^2}}{{x_0^2}}x - \frac{{v_0^2}}{{{x_0}}}.....(ii)\,\,\left[ {\because a = v\frac{{dv}}{{dx}}} \right] \cr} $$
On comparing the equation (ii) with equation of a straight line
$$\eqalign{ & y = mx + c \cr & {\text{we get }}m = \frac{{v_0^2}}{{x_0^2}} = + ve, \cr & i.e.,\,\tan \theta = + ve,\,\,\,i.e.,\,\,\theta \,\,{\text{is acute}}{\text{.}} \cr & {\text{Also}}\,\,c = - \frac{{v_0^2}}{{x_0^2}}, \cr} $$
$$i.e.,$$ the $$y$$-intercept is negative
The above conditions are satisfied in graph (A).
