A body of mass 2 text kg$2 \text{ kg}$ is moving along x-direction such that its displacement as function of time is given by x(t) = alpha t^2 + beta t + gamma text m$x(t) = \alpha t^{2} + \beta t + \gamma \text{ m}$, where alpha = 1 text m/s^2$\alpha = 1 \text{ m/s}^{2}$, beta = 1 text m/s$\beta = 1 \text{ m/s}$ and gamma = 1 text m$\gamma = 1 \text{ m}$. The work done on the body during the time interval t = 2 text s$t = 2 \text{ s}$ to t = 3 text s$t = 3 \text{ s}$, is ________ J.

### Related Formula W = int F cdot dx = F cdot Delta x quad (textif force is constant)$$W = \int F \cdot dx = F \cdot \Delta x \quad (\text{if force is constant})$$ v = fracdxdt, quad a = fracdvdt$$v = \frac{dx}{dt}, \quad a = \frac{dv}{dt}$$ F = m a$F = m a$ ### Core Logic Given displacement equation: x(t) = 1t^2 + 1t + 1$$x(t) = 1t^2 + 1t + 1$$ Velocity is the first derivative: v(t) = 2t + 1$v(t) = 2t + 1$ Acceleration is the second derivative: a(t) = 2 text m/s^2$$a(t) = 2 \text{ m/s}^2$$ Since acceleration is constant, the force is also constant: F = m a = 2 times 2 = 4 text N$$F = m a = 2 \times 2 = 4 \text{ N}$$ ### Step 1: Finding Displacement Interval We need the displacement during the interval t=2$t=2$ to t=3$t=3$: x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 text m$$x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 \text{ m}$$ x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 text m$$x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 \text{ m}$$ Total displacement S = x(3) - x(2) = 13 - 7 = 6 text m$S = x(3) - x(2) = 13 - 7 = 6 \text{ m}$ ### Step 2: Final Conclusion Work done: W = F cdot S = 4 text N times 6 text m = 24 text J$$W = F \cdot S = 4 \text{ N} \times 6 \text{ m} = 24 \text{ J}$$ ### Pattern Recognition If x(t)$x(t)$ is a quadratic in t$t$, acceleration is constant. Work done simply becomes ma times Delta x$ma \times \Delta x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics Class 11 Physics: Work, Energy and Power