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Grade: 11

                        

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.

2 years ago

Answers : (1)

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