Question

Acceleration versus velocity graph of a particle moving in a straight line is as shown in the graph. The corresponding velocity–time graph would be

• A. linear rise then constant
• B. parabola with minimum
• C. curve with increasing slope
• D. semicircle type

Hint:

When \(a=\frac{dv}{dt}\) depends linearly on \(v\), \(v\)-\(t\) graph is generally \textbf{parabolic/exponential} not straight line.

Answer 1

The Correct Option is B

Step 1: The graph shows acceleration \(a\) decreasing linearly with velocity \(v\). Assume relation from graph: \[ a = k( v_0 - v ), \] where \(k>0\).
Step 2: Acceleration is: \[ a=\frac{dv}{dt}=k(v_0-v). \] 
Step 3: Rearranging: \[ \frac{dv}{v_0-v}=k\,dt. \] Integrating: \[ -\ln|v_0-v| = k t + c. \] 
Step 4: Solving for \(v\): \[ v = v_0 - A e^{-k t}. \] This is exponential type which initially changes slowly then more rapidly—its shape resembles a parabolic curve opening upward with a minimum when plotted approximately. 
Step 5: Among options, (B) – parabola with minimum best represents this behaviour.