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

A body covers 26,28,30 and 32 meters in and seconds respectively. The body start

A body covers 26,28,30 and 32 meters in \( 10^{t h}, 11^{t h}, 12^{t h} \) and \( 13^{t h} \) seconds respectively. The body starts

  • A. from rest and moves with uniform velocity
  • B. from rest and moves with uniform acceleration
  • C. with an initial velocity and moves with uniform acceleration
  • D. with an initial velocity and moves with uniform velocity

Step-by-step solution

The distances covered in successive seconds increase by a constant amount (2 m), indicating uniform acceleration. Using the formula for distance in nth second, s_n = u + a/2 (2n-1), for the 10th second: 26 = u + a/2 (19). The difference between consecutive seconds gives a = 2 m/s². Solving gives u = 7 m/s, so initial velocity is non-zero. Hence, the body starts with an initial velocity and moves with uniform acceleration.
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