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NCERT Exemplar · Q36

Q.A body of unit mass (1 kg1\ \text{kg}) moves in a plane, described by two velocity–time graphs. The xx-velocity graph is triangular: vxv_x rises linearly from 00 at t=0t=0 to 2 m s−12\ \text{m s}^{-1} at t=1 st=1\ \text{s}, then falls linearly back to 00 at t=2 st=2\ \text{s}, and remains 00 for t>2 st>2\ \text{s}. The yy-velocity graph rises linearly from 00 at t=0t=0 to 1 m s−11\ \text{m s}^{-1} at t=1 st=1\ \text{s}, and then stays constant at 1 m s−11\ \text{m s}^{-1} for t>1 st>1\ \text{s}. Find the force acting on the body as a function of time.

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Acceleration is the slope of a velocity–time graph, and with m=1 kgm=1\ \text{kg} the force numerically equals the acceleration. Reading the slopes piecewise gives a force that changes in three time intervals.

Concept

F=ma\mathbf F=m\mathbf a with m=1 kgm=1\ \text{kg}, and ax=dvxdta_x=\dfrac{dv_x}{dt}, ay=dvydta_y=\dfrac{dv_y}{dt} are the slopes of the two graphs.

xx-component (triangular vxv_x)

  • 0<t<1 s0<t<1\ \text{s}: vxv_x goes 0→20\to2, slope ax=+2 m s−2⇒Fx=+2 Na_x=+2\ \text{m s}^{-2}\Rightarrow F_x=+2\ \text{N}.
  • 1<t<2 s1<t<2\ \text{s}: vxv_x goes 2→02\to0, slope ax=−2 m s−2⇒Fx=−2 Na_x=-2\ \text{m s}^{-2}\Rightarrow F_x=-2\ \text{N}.
  • t>2 st>2\ \text{s}: vx=0v_x=0, so Fx=0F_x=0.

yy-component

  • 0<t<1 s0<t<1\ \text{s}: vyv_y goes 0→10\to1, slope ay=+1 m s−2⇒Fy=+1 Na_y=+1\ \text{m s}^{-2}\Rightarrow F_y=+1\ \text{N}.
  • t>1 st>1\ \text{s}: vy=1 m s−1v_y=1\ \text{m s}^{-1} constant, so Fy=0F_y=0.

Force as a function of time …

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