JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

The velocity ( $v$ ) – Distance ( $x$ ) graph is shown in figure. Which graph represents acceleration ( $a$ ) versus distance ( $x$ ) variation of this system?

Solution

From the linear $v$ - $x$ graph: $v = C_{1} - C_{2} x$ where $C_{1}, C_{2} > 0$ .

Acceleration: $a = v \frac{d v}{d x} = (C_{1} - C_{2} x) (- C_{2}) = - C_{1} C_{2} + C_{2}^{2} x$ .

This is linear in $x$ with positive slope ( $C_{2}^{2}$ ) and negative intercept ( $- C_{1} C_{2}$ ). Crosses the $x$ -axis at $x = C_{1} / C_{2}$ .

The graph in option (2) matches: straight line, positive slope, negative $a$ -intercept.

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