Question 58: Motion in 1D

Question

The velocity $-$ displacement graph of a particle is shown in the figure.
The acceleration $-$ displacement graph of the same particle is represented by :

Accepted Answer

The slope of the given v versus x graph is $m =  - {{{v_0}} \over {{x_0}}}$ and intercept is c = + v
0
. Hence, v varies with x as
$v =  - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$ ...... (1)
where v
0
and x
0
are constants of motion. Differentiating with respect to time t, we have
${{dv} \over {dt}} =  - \left( {{{{v_0}} \over {{x_0}}}} \right){{dx} \over {dt}}$
or $a =  - \left( {{{{v_0}} \over {{x_0}}}} \right)v$ ........ (2)
Using Eq. (1) in Eq. (2), we get
$a =  - \left( {{{{v_0}} \over {{x_0}}}} \right)\left( { - {{{v_0}} \over {{x_0}}}x + {v_0}} \right)$
$a = {\left( {{{{v_0}} \over {{x_0}}}} \right)^2}x - {{v_0^2} \over {{x_0}}}$
Thus the graph of a versus x is a straight line having a positive slope = ${\left( {{{{v_0}} \over {{x_0}}}} \right)^2}$ and negative intercept = $ - {{v_0^2} \over {{x_0}}}$. Hence the correct choice is (a).

Answer

Answer

Answer

Answer

Motion in 1D

The velocity - displacement graph of a particle is shown in the figure. The acceleration - displacement graph of the same particle is represented by :

Choose the correct answer: