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The given graph shows the variation of velocity (v) with position (x) for a particle moving along
a straight line
Which of the following graph shows the variation of acceleration (a) with position (x)?
Options:
a straight line
Which of the following graph shows the variation of acceleration (a) with position (x)?
- A
- B
- C
- D
Solution:
1295 Upvotes
Verified Answer
The correct answer is:
(A)
Given line have positive intercept but negative slope.
So its equation can be written as,
\( v=-m x+v_{0} \ldots(1) \)
where \( m=\tan \theta=\frac{v_{0}}{x_{0}} \)
By differentiating with respect to time we get, \( \frac{d v}{d t}=-m \frac{d x}{d t}=-m v \)
Now substituting the value of ' \( v \) ' from equation (1) we get
\( \frac{d v}{d x}=-m\left[-m x+v_{0}\right]=m^{2} x-m v_{0} \) \( a=m^{2} x-m v_{0} \)
i.e., the graph between a and \( x \) should have positive slope but negative intercept on acceleration axis.
Hence option (A) is correct.
Given line have positive intercept but negative slope.
So its equation can be written as,
\( v=-m x+v_{0} \ldots(1) \)
where \( m=\tan \theta=\frac{v_{0}}{x_{0}} \)
By differentiating with respect to time we get, \( \frac{d v}{d t}=-m \frac{d x}{d t}=-m v \)
Now substituting the value of ' \( v \) ' from equation (1) we get
\( \frac{d v}{d x}=-m\left[-m x+v_{0}\right]=m^{2} x-m v_{0} \) \( a=m^{2} x-m v_{0} \)
i.e., the graph between a and \( x \) should have positive slope but negative intercept on acceleration axis.
Hence option (A) is correct.
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