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consider a particle initially moving with 5 m s-1 starts decelerating at a constant rate of 2 m s -2. find the distance travelled in the third second.

Purva Agrawal , 6 Years ago
Grade 11
anser 1 Answers
Khimraj

Last Activity: 6 Years ago

velocity after 2 sec = 5 – 2*2 = 1m/s
so distance covered in next 1/2 sec = 1*1 – ½*2*(1/2)2 = ¾ m
now its velocity become zero and reverse the direction
then distance in next ½ sec = ½*2*(1/2)2 = ¼ M
SO distance in 3rd sec = ¾ + ¼ = 1m.

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