how to derivea relation for the distance travelled in nth second by a uniformly accelerated body

distance travelled in n seconds:

S1= un + 1/2a(n^2)

distance travelled in (n-1) seconds:

S2 = u(n-1) + 1/2a((n-1)^2)

distance travelled in nth second:

S1-S2

= u + 1/2a(2n-1)

actually the other given answer is also correct.

displacement in the nth second,we r talking about.suppose accleration is a and the initial velocity is u.

s=ut+1/2at^2.

if i will be said to find out the displacement in the nth second.so i will find out the displacement in the n-1 th second,can be easily found by the above given equation.

s(n-1)=u(n-1)+1/2a(n-1)^2=un-u+1/2a(n^2-2n+1)=un-u+1/2an^2-an+1/2a

now this is the total displacement in n-1 second.in n second the displacent is :

s(n)=un+1/2an^2

so i f i want the displacement in that n second nly,so i need to subtract the total displacement in n second and in n-1 second.

s(n)-s(n-1)=un+1/2an^2-un+u-1/2an^2+an-1/2a=u+an-1/2a=u+1/2a(2n-1).

actually to make this answer correct u need to write the seconds (dimensionally correct)

so,u(1s)+1/2a(2n-1s)(1s) is the correct answer.if u will keep on writing the units from the first then u will get this equation only.

Distance travelled by a uniformly accelerated body in nth second can be obtained by subtracting the distance travelled by the body in (n-1) second from the distance travelled by the body in n seconds.If, u= initial velocity of the body;a= uniform acceleration of the body;Sn,Sn-1= distances travelled by the body in n seconds and (n-1) seconds respectively,Dn = distance travelled in nth second.Then, Dn = Sn-Sn-1We know that the distance travelled in t seconds is given by; S= ut+1/2 at×t=> distance travelled in n seconds is, Sn= un+1/2an×nDistance travelled in (n-1) seconds is, Sn-1= u (n-1)+1/2a (n-1)(n-1)Now putting these values in, Dn=Sn-Sn-1We get, Dn=[un+1/2an×n]-[u (n-1)+1/2a (n-1)(n-1)] =un+1/2an×n-un+u-1/2an×n+an-a/2 =u+an-a/2 =u+a (n-1/2) or Dn= u+a/2 (2n-1)

Home » Forum » Modern Physics » kinematicshow to derivea relation for the distance travelled in nth second by a uniformly accelerated body7 years agoAnswers : (3)distance travelled in n seconds:S1= un + 1/2a(n^2)distance travelled in (n-1) seconds:S2 = u(n-1) + 1/2a((n-1)^2)distance travelled in nth second:S1-S2= u + 1/2a(2n-1)

Dear Student,
Please find below the solution to your problem.

displacement in the nth second,we r talking about.suppose accleration is a and the initial velocity is u.

s=ut+1/2at^2.

if i will be said to find out the displacement in the nth second.so i will find out the displacement in the n-1 th second,can be easily found by the above given equation.

s(n-1)=u(n-1)+1/2a(n-1)^2=un-u+1/2a(n^2-2n+1)=un-u+1/2an^2-an+1/2a

now this is the total displacement in n-1 second.in n second the displacent is :

s(n)=un+1/2an^2

so i f i want the displacement in that nsecond nly,so i need to subtract the total displacement in n second and in n-1 second.

s(n)-s(n-1)=un+1/2an^2-un+u-1/2an^2+an-1/2a=u+an-1/2a=u+1/2a(2n-1).

actually to make this answer correct u need to write the seconds (dimensionally correct)

so,u(1s)+1/2a(2n-1s)(1s) is the correct answer.if u will keep on writing the units from the first then u will get this equation only.

Thanks and Regards