A particle has an initial velocity of 9 m / s due east and a constant acceleration of 2 m / s 2 due west. The distance covered by the particle in the fifth second of its motion is:

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TestHub · Accepted Answer

The particle stops at t = 4.5 s, midway through the fifth second. At t = 4 s its speed is 9 − 2 ( 4 ) = 1 m/s, so it covers 2 1 ​ ( 0.5 ) ( 1 ) = 0.25 m before stopping and another 0.25 m moving back, a total of 0.5 m. Option B.

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