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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line5 Marks
Question
The velocity$-$displacement graph of a particle is shown in Fig.
  1. Write the relation between $v$ and $x$.
  2. Obtain the relation between acceleration and displacement and plot it.
✓

Answer

In this problem, we will use equation of straight line graph $($linear equation$). y = mx + c$. In this equation, $m$ is the slope of the graph and $c$ is the interception on $y-$axis.

 Now according to the problem, initial velocity $= v_0$ Let the distance travelled in time $t = x_0$. For the graph $\tan\theta=\frac{\text{v}_0}{\text{x}_01}=\frac{\text{v}_0-\text{v}}{\text{x}}\ \ ...(\text{i})$ where, $v$ is velocity and $x$ is displacement at any instant of time $t.$ From Eq. $(i), \text{v}_0-\text{v}=\frac{\text{v}_0}{\text{x}_0}\text{x} $
$\Rightarrow\text{v}=\frac{-\text{v}_0}{\text{x}_0}\text{x}+\text{v}_0$ We know that Acceleration $\text{a}=\frac{\text{dv}}{\text{dt}}=\frac{-\text{v}_0}{\text{x}_0}\frac{\text{dx}}{\text{dt}}+0$
$\Rightarrow\text{a}=\frac{-\text{v}_0}{\text{x}_0}(\text{v})$ $=\frac{-\text{v}_0}{\text{x}_0}\Big(\frac{-\text{v}_0}{\text{x}_0}\text{x}+\text{v}_0\Big)=\frac{\text{v}_0^2}{\text{x}_0^2}\text{x}-\frac{\text{v}_0^2}{\text{x}_0}$ Graph of a versus $x$ is given below.
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