Question

The velocity displacement graph of a particle moving along a straight line is shown in figure.

Then the acceleration displacement graph is.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

As per the question from the graph, we can write
$\upsilon =-\text{x}+2$
$\text{a}=\frac{\text{dv}}{\text{dt}}=-\frac{\text{dx}}{\text{dt}}=-\upsilon$
$\upsilon =-(-\text{x}+2)=\text{x}-2$
Therefore,
At $\text{x}=0\text{m}$ we get,
$\upsilon =0-2=-2\text{m/s}$
At $\text{x}=1m$ we get,
$\upsilon =1-2=-1\frac{\text{m}}{\text{s}}$
At $\text{x}=2\text{m}$ we get
$\upsilon =2-2=0\text{m/s}$
The velocity $–$ displacement graph will be,

According to above calculations only (c) option justifies the equation.

Final Answer: (c)