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Physics · Uniformly accelerated motion, velocity-time, position-time graph, relations for uniformly accelerated motion, relative velocity

A block is thrown with a velocity of (relative to ground) on a belt, which is mo

A block is thrown with a velocity of \( 2 m / s \) (relative to ground) on a belt, which is moving with velocity \( 4 m / s \) in opposite direction of the initial velocity of block. If the block stops slipping on the belt after \( 4 s \) of the throwing then choose the correct statement(s) This question has multiple correct options

  • A. Displacement with respect to ground is zero after 2.66 and displacement with respect to ground is \( 12 \mathrm{m} \) after \( 4 s \)
  • B. Displacement with respect to ground in \( 4 s \) is \( 4 m \).
  • C. Displacement with respect to belt in \( 4 s \) is \( -12 m \).
  • D. Displacement with respect to ground is zero in \( \frac{8}{3} s \)

Step-by-step solution

The block starts with velocity +2 m/s relative to ground, and the belt moves with -4 m/s relative to ground. The block stops slipping after 4 s, meaning its velocity becomes -4 m/s. Acceleration is constant: a = (v - u)/t = (-4 - 2)/4 = -1.5 m/s². Displacement relative to ground: s = ut + ½at². At t = 8/3 s, s = 2*(8/3) + 0.5*(-1.5)*(64/9) = 16/3 - (0.75*64/9) = 16/3 - 48/9 = 48/9 - 48/9 = 0. At t=4 s, s = 2*4 + 0.5*(-1.5)*16 = 8 - 12 = -4 m, not 12 m. Displacement relative to belt: initial relative velocity = 2 - (-4) = 6 m/s, final relative velocity = 0, so relative acceleration = -1.5 m/s², relative displacement = 6*4 + 0.5*(-1.5)*16 = 24 - 12 = 12 m, not -12 m. Only option D is correct.
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