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Grade 11Trigonometry

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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.

Profile image of Purva Agrawal
7 Years agoGrade 11
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1 Answer

Profile image of Arun
7 Years ago
 
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.

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