Question 9: Motion in 1D

Question

The velocity $(v)$ - Distance $(x)$ graph is shown in figure. Which graph represents acceleration(a) versus distance ( $x$ ) variation of this system?

Accepted Answer

For the given graph shows a straight line with a negative slope and a positive y -intercept.
We can represent this line using the standard equation for a straight line, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ :
$$ \mathrm{v}=-\mathrm{mx}+\mathrm{v}_0 $$
Where:
1. $-m$ is the constant negative slope $(m>0)$.
2. $\mathrm{v}_0$ is the initial velocity at $\mathrm{x}=0$ (the y -intercept).
Acceleration a can be expressed in terms of velocity v and position x using the following formula :
$$ a=v \frac{d v}{d x} $$
$$\Rightarrow $$ $\mathrm{a}=\left(-\mathrm{mx}+\mathrm{v}_0\right) \frac{\mathrm{d}}{\mathrm{dx}}\left[-\mathrm{mx}+\mathrm{v}_0\right]$
$$\Rightarrow $$ $\mathrm{a}=\left(-\mathrm{mx}+\mathrm{v}_0\right)[-\mathrm{m}]$
$$\Rightarrow $$ $$ \mathrm{a}=\mathrm{m}^2 \mathrm{x}-\mathrm{mv}_0 $$
The equation $\mathrm{a}=\mathrm{m}^2 \mathrm{x}-\mathrm{mv}_0$ (linear function of x ) is also the equation of a straight line ( $\mathrm{y}=\mathrm{MX}+\mathrm{C}$ ) where,
Slope, $\mathrm{M}=\mathrm{m}^2>0$. So the slope is positive.
Intercept, $\mathrm{C}=-\mathrm{mv}_0$. Since both m and $\mathrm{v}_0$ are positive values from our initial graph, the intercept is negative.
So, the resulting graph is,
Therefore, the correct graph representing the acceleration versus distance variation is option D.
Hence, the correct option is (D).

Answer

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Answer

Motion in 1D

The velocity (v) - Distance (x) graph is shown in figure. Which graph represents acceleration(a) versus distance ( x ) variation of this system?

Choose the correct answer: