Question 2: Motion in 1D

Question

What will be the a vs x graph for the following
graph?
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1m59dt66a/ca2cbb82-a8f9-4851-a101-3d785eb7b247/8c408d30-c5c3-11ef-9ad1-6d283e3c94c4/file-jaoe38c1m59dt66b.png?format=png)

Accepted Answer

Equation of given line when velocity is plotted against position with intercept $=v_0$ and slope $=v_0 / x_0$ is,
$v=-\frac{v_0 x}{x_0}+v_0$ ...... (i)
($-$Ve sign is became the slope is decreasing) On differentiating with respect to $t$
$\frac{d v}{d t}=-\frac{v_0}{x_0} \frac{d x}{d t}+0$
or $\quad a=-\frac{v_0}{x_0} v$ ...... (ii)(
Putting $v$ from Eq. (i) in Eq. (ii),
$\begin{aligned}
a & =-\frac{v_0}{x_0}\left[-\frac{v_0}{x_0} x+v_0ight] \
& =\frac{v_0^2}{x_0^2} x-\frac{v_0^2}{x_0}
\end{aligned}$
Which represents a straight line with positive slope and negative intercept.

Answer

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Answer

Motion in 1D

What will be the a vs x graph for the following graph?

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