Question

Acceleration (\(a\)) – displacement (\(s\)) graph of a particle moving in a straight line is as shown in the figure. The initial velocity of the particle is zero. The \(v\)-\(s\) graph of the particle would be:

• A. Option a
• B. Option b
• C. Option c
• D. Option d

Hint:

Useful relation: \[ a = v\frac{dv}{ds} \]
If \(a \propto s\), then \(v \propto s\)
Shape of \(v\)-\(s\) graph follows directly from integration

Answer 1

The Correct Option is D

Step 1: Interpret the given \(a\)-\(s\) graph. From the graph:
Acceleration increases linearly with displacement.
Hence, \[ a \propto s \quad \Rightarrow \quad a = ks \] where \(k\) is a positive constant.
Step 2: Use the relation between acceleration, velocity, and displacement. For one–dimensional motion: \[ a = v\frac{dv}{ds} \] Substitute \(a = ks\): \[ v\frac{dv}{ds} = ks \]
Step 3: Integrate. \[ v\,dv = ks\,ds \] \[ \int v\,dv = k \int s\,ds \] \[ \frac{v^2}{2} = \frac{k s^2}{2} + C \]
Step 4: Apply initial condition. Given initial velocity is zero at \(s=0\): \[ v=0 \Rightarrow C=0 \] So, \[ v^2 = ks^2 \] \[ v = \sqrt{k}\,s \]
Step 5: Deduce the nature of the \(v\)-\(s\) graph.
Velocity is directly proportional to displacement.
Hence, the \(v\)-\(s\) graph is a straight line passing through the origin. Final Answer: \[ \boxed{\text{Option (d)}} \]