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

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2 years ago

Arun
24742 Points
```							 V=U+ATSince v=0, we substitute values.0=5-2t(because it is decelerating and hence it is negative).T=2.5sDisplacement in nth second is given asSn=u+a(2n-1)/2For displacement in second second, we get S=5-1(3)=12For displacement in third second, we get S=5-1(5)=20.
```
2 years ago
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