JEE PYQ: Motion in a Straight Line - Question ID f99ebda1ea55 (JEE Main 2021)

In kinematics, the displacement \( x(t) \) of a particle moving along a straight line is obtained by integrating its velocity \( v(t) \) with respect to time. The fundamental relation is: $x(t) \;=\; \int_{0}^{t} v(t')\,dt'.$ Given the velocity function $v(t) \;=\; v_{0} + g\,t + F\,t^{2},$ we integrate term by term from \( t=0 \) to \( t=1 \) to find the displacement at \( t=1 \).

1. Write the integral for displacement: $x(1) \;=\; \int_{0}^{1} \bigl(v_{0} + g\,t + F\,t^{2}\bigr)\,dt.$ 2. Integrate each term separately: - Constant term: $\int_{0}^{1} v_{0}\,dt = v_{0}\int_{0}^{1} dt = v_{0}\cdot(1-0) = v_{0}.$ - Linear term in \( t \): $\int_{0}^{1} g\,t\,dt = g\int_{0}^{1} t\,dt = g\,\Bigl[\tfrac{t^{2}}{2}\Bigr]_{0}^{1} = g\cdot\tfrac{1}{2}.$ - Quadratic term in \( t \): $\int_{0}^{1} F\,t^{2}\,dt = F\int_{0}^{1} t^{2}\,dt = F\,\Bigl[\tfrac{t^{3}}{3}\Bigr]_{0}^{1} = F\cdot\tfrac{1}{3}.$ 3. Sum the results: $x(1) = v_{0} + \frac{g}{2} + \frac{F}{3}.$ 4. Compare with the given options. This matches option B.

• Students often confuse velocity with displacement and simply plug \( t=1 \) into \( v(t) \), leading to option A. • Forgetting to divide by the new power when integrating \( t \) and \( t^{2} \) can produce incorrect coefficients. • Always check that the integration constants vanish when evaluating definite integrals from \( 0 \) to \( t \).