# sir,how to calculate the distance travelled by a body in 1d motion or freely falling motion during the last second of its motion..and 1 question is there in 1d motionu=9m/s due east, a=2 m/s2 due west.find the distance travelled by the body during its 5th second of motion

509 Points
13 years ago

dear dashmesh , here is the simple method to solve such type of problems ....

distance travelled in n seconds (Sn) = gn2/2

distance travelled in (n-1) seconds (Sn-1) = g(n-1)2/2

distance travelled in nth second =  Sn - Sn-1

Snth = g/2 [ n2 - (n-1)2 ]

= g/2 [ (2n-1) ]  .......................1

this is the formula for finding distance travelled in nth second....

distance travelled during fifth second is S5th at n = 5

S5th = g/2[ (2*5-1)] = 9g/2                 (g  = 10m/s2)

=45m

approve my ans if u like it

Himanshu Dogra
34 Points
13 years ago

Here's a formula for the distance traversed by the body in nth second:

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

u = intial velocity of the body,

a = acceleration of the body (with proper algebric signs)

you can also apply the same formula for displacement by using vector addition and substraction formula.

rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method
rajeev kumar
8 Points
10 years ago

Snth = u + a/2 (2n-1)

by calculas method