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Distance travelled by particle starting from rest and moving with an acceleration of 4/3 m/s2 in nth second will be??

Distance travelled by particle starting from rest and moving with an acceleration of 4/3 m/s2 in nth second will be??

Grade:12th pass

5 Answers

Aman gohel
71 Points
6 years ago
S(in nth sec)=u+a/2(2n-1). =>0+4/(3*2)(2n-1). =>Therefore the distance travelled by the particle in nth second is:-. =>2/3(2n-1). [Ans]
Vikas TU
14149 Points
6 years ago
Dear Student,
Given data:
u=0, a=4/3 m/s2
velocity at the start of nth second = u+at
=>v=4(n-1)/3
distance travelled in nth second = S= ut+at2/2
=>S=4(n-1)/3 + 4/6
=>S=(4n-2)/3 m
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Nitesh
102 Points
6 years ago
From the equation of motion we get the formula for distance travelled in nth second is Snth=u+(2n-1)×a×1/2.putting the value a=4/3 m/s2,u=0 because starts from rest.We get Snth=(2n-1)×2/3.Hence distance travelled in nth second is 2 (2n-1)/3
Rahul Krishna
38 Points
6 years ago
distance travelled by nth second=u+a/2*(2*n-1))........here u=0....a=4/3.....so..........s=4/3*1/2*(2*n-1)=4/6*(2*n-1)={\color{Red} 2/3*(2*n-1).}....correct answer.......
yashveer
27 Points
5 years ago
 
Dear Student,
Given data:
u=0, a=4/3 m/s2
velocity at the start of nth second = u+at
=>v=4(n-1)/3
distance travelled in nth second = S= ut+at2/2
=>S=4(n-1)/3 + 4/6
=>S=(4n-2)/3 m
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
 

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