Q.44. A body of mass 2 kg is moving along x-direction such that its displacement as function of time

Q: What is the correct answer for JEE Main Physics Q44?

The correct answer is C. Mass of the body $m = 2$ kg.Displacement $x(t) = \alpha t^2 + \beta t + \gamma$.Given $\alpha = 1 \text{ m/s}^2$, $\beta = 1 \text{ m/s}$, $\gamma = 1\text{m}$.So, $x(t) = t^2 + t + 1$.Velocity $v(t) = \frac{dx}{dt} = 2t + 1$.Acceleration $a(t) = \frac{dv}{dt} = 2$.Force $F = ma = 2 \text{ kg} \times 2 \text{ m/s}^2 = 4 \text{ N}$.Since the force is constant, we can calculate work done as $W = F \

Q: Q.44. A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by , where , and . The work done on the body during the time interval to , is…

The correct answer is C. Mass of the body $m = 2$ kg.Displacement $x(t) = \alpha t^2 + \beta t + \gamma$.Given $\alpha = 1 \text{ m/s}^2$, $\beta = 1 \text{ m/s}$, $\gamma = 1\text{m}$.So, $x(t) = t^2 + t + 1$.Velocity $v(t) = \frac{dx}{dt} = 2t + 1$.Acceleration $a(t) = \frac{dv}{dt} = 2$.Force $F = ma = 2 \text{ kg} \times 2 \text{ m/s}^2 = 4 \text{ N}$.Since the force is constant, we can calculate work done as $W = F \

Q: What are the options for this JEE Main Physics question?

A. 49 · B. 42 · C. 24 · D. 12

Question

Q.44. A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by , where ,  and . The work done on the body during the time interval  to , is .......... J.

FirstInTest · Accepted Answer

Mass of the body $m = 2$ kg.Displacement $x(t) = \alpha t^2 + \beta t + \gamma$.Given $\alpha = 1 	ext{ m/s}^2$, $\beta = 1 	ext{ m/s}$, $\gamma = 1	ext{m}$.So, $x(t) = t^2 + t + 1$.Velocity $v(t) = \frac{dx}{dt} = 2t + 1$.Acceleration $a(t) = \frac{dv}{dt} = 2$.Force $F = ma = 2 	ext{ kg} 	imes 2 	ext{ m/s}^2 = 4 	ext{ N}$.Since the force is constant, we can calculate work done as $W = F 	imes \Delta x$.Displacement at $t=2$ s: $x(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7$ m.Displacement at $t=3$ s: $x(3) = (3)^2 + 3 + 1 = 9 + 3 + 1 = 13$ m.Change in displacement $\Delta x = x(3) - x(2) = 13 - 7 = 6$ m.Work done $W = F 	imes \Delta x = 4 	ext{ N} 	imes 6 	ext{ m} = 24 	ext{ J}$.Alternatively, using Work-Energy Theorem: $W = \Delta K = K_f - K_i$.Velocity at $t=2$ s: $v(2) = 2(2) + 1 = 5 	ext{ m/s}$.Velocity at $t=3$ s: $v(3) = 2(3) + 1 = 7 	ext{ m/s}$.Initial kinetic energy $K_i = \frac{1}{2}mv(2)^2 = \frac{1}{2}(2)(5)^2 = 25 	ext{ J}$.Final kinetic energy $K_f = \frac{1}{2}mv(3)^2 = \frac{1}{2}(2)(7)^2 = 49 	ext{ J}$.Work done $W = K_f - K_i = 49 - 25 = 24 	ext{ J}$.

Answer

49

Answer

42

Answer

24

Answer

12

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