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```
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 motion u=9m/s due east, a=2 m/s2 due west. find the distance travelled by the body during its 5th second of motion

```
9 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
```
9 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.
```
9 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
```							 Snth = u + a/2 (2n-1)by calculas method
```
6 years ago
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