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# A particle starts from rest has an acceleration of a =2tm/s^2 then distance travelled in first second

Grade:11

## 4 Answers

Mohammed Imran
28 Points
3 years ago
Initial velocity is zero And question asked is distance traveled in 1st secondS=u + a/2(2n-1)U =0a=2m/s^2n=1So put this in formula Answer would be 1m
Arun
25763 Points
3 years ago
Given that
a=2t
now we know that
dv / dt = 2t
dv = 2t dt
integrate it
v = t2
now we know that
ds/ dt = v
ds  = t2 dt
integrate it
s = t3/3
hence in first second
put t = 1 second
hence
s = 1/3 metre
hope it helps
Shailendra Kumar Sharma
188 Points
3 years ago
Acceleration, (a) = 2tv= 2t^2/2 + constant= t2@ t= 0 v= 0So could constant term is zero V= t2Now, X = t^3/3 +cAt t= 0, X= 0So X = T3/3So distance covered in 1 sec X=1/3
Lakshay Kumar
43 Points
3 years ago
Since acceleration of body is not constant we have to integrate it two times and the result will be x=1/3 metres, where x is displacement

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