Velocity–Position Graph: Find the Variation of Acceleration with Position | JEE Main Physics
Questions involving velocity-position (v-x) graphs are frequently asked in JEE Main. Such problems require the application of the relation between velocity, acceleration, and displacement rather than direct differentiation with respect to time.
In this problem, the velocity of a particle decreases linearly with position. We have to determine how acceleration varies with position.
Question
The velocity of a particle varies with position as shown in the graph. Find the correct equation representing the variation of acceleration with position.
Concept Used
The relation between acceleration and velocity when velocity is given as a function of position is:
a = v × (dv/dx)
This formula is obtained using the chain rule:
a = dv/dt
= (dv/dx)(dx/dt)
= v(dv/dx)
Step 1: Write the Equation of the Velocity-Position Graph
The given graph is a straight line having:
- Negative slope
- Positive y-intercept
Hence, the equation of the graph is:
v = -mx + c
where,
- m = magnitude of slope
- c = velocity intercept
Step 2: Differentiate Velocity with Respect to Position
Differentiate the velocity equation:
v = -mx + c
dv/dx = -m
Since the graph is linear, the slope remains constant.
Step 3: Apply the Acceleration Formula
Using
a = v(dv/dx)
a = (-mx + c)(-m)
a = m²x - mc
Final Equation
a = m²x - mc
Nature of the Acceleration–Position Graph
The obtained equation is of the standard straight-line form:
y = Mx + C
Comparing,
- Slope = m² (Positive)
- Y-intercept = -mc (Negative)
Therefore, the acceleration-position graph is:
- A straight line
- Positive slope
- Negative y-intercept
Graph Interpretation
- Velocity decreases uniformly with position.
- Since dv/dx is constant, acceleration depends linearly on position.
- Acceleration increases uniformly as the particle moves forward.
- The acceleration graph is a straight line rising from a negative intercept.
Key Formula to Remember
| Situation | Formula |
|---|---|
| Acceleration from velocity-position graph | a = v(dv/dx) |
| Velocity equation of straight line | v = mx + c |
| Standard line equation | y = mx + c |
Important JEE Concepts Tested
- Chain Rule in differentiation
- Relationship between velocity and acceleration
- Slope of a graph
- Straight-line equations
- Kinematics using graphical interpretation
- Velocity as a function of position
Common Mistakes Students Make
- Using a = dv/dt instead of a = v(dv/dx).
- Ignoring the negative slope of the velocity graph.
- Differentiating with respect to time instead of position.
- Missing the negative sign while calculating dv/dx.
- Drawing the acceleration graph with a negative slope instead of a positive slope.
Quick Exam Trick
- If the v-x graph is a straight line, then dv/dx is constant.
- Multiply velocity by the constant slope to obtain acceleration.
- If the velocity graph has a negative slope, acceleration may still have a positive slope after multiplication depending on the sign of velocity.
- Always express the final equation in the form a = Ax + B to identify the graph quickly.
Practice Questions
- If v = 8 − 2x, find the acceleration as a function of position.
- A particle has velocity v = 12 − 3x. Determine the acceleration when x = 2 m.
- For v = 5x + 4, obtain the equation of acceleration with position.
- If velocity decreases linearly with displacement, explain the nature of the acceleration-position graph.
Final Answer
Using the relation a = v(dv/dx) and the velocity equation v = -mx + c, the acceleration is:
a = m²x - mc
Hence, the acceleration varies linearly with position and its graph is a straight line having a positive slope (m²) and a negative y-intercept (-mc).