Question

The area of the acceleration–displacement curve of a body gives:

• A. Impulse
• B. Change in momentum per unit mass
• C. Change in kinetic energy per unit mass
• D. Velocity

Hint:

Graphical interpretations:
Area under \(v\)-\(t\) curve → displacement
Area under \(a\)-\(t\) curve → change in velocity
Area under \(a\)-\(s\) curve → change in KE per unit mass

Answer 1

The Correct Option is C

Step 1: Recall the relation between acceleration, velocity, and displacement. \[ a = v\frac{dv}{ds} \]
Step 2: Rearrange the equation. \[ a\,ds = v\,dv \]
Step 3: Integrate both sides. \[ \int a\,ds = \int v\,dv \] \[ \text{Area under } a\text{–}s \text{ curve} = \frac{v^2}{2} \]
Step 4: Interpret the result. \[ \frac{v^2}{2} = \text{kinetic energy per unit mass} \] Hence, the area gives the change in kinetic energy per unit mass. Final Answer: \[ \boxed{\text{Change in kinetic energy per unit mass}} \]