Find the ratio of distance travelled by the free-falling body in 1st, 2nd and 3rd second: {Ans S₁ : S₂ : S₃ = 1: 3: 5}
Given Data:
t₁ = 1 s
t₂ = 2 s
t₃ = 3 s
To Find:
S₁ : S₂ : S₃ = ?
Solution:
Let
S₁ =distance travelled in 1st second
S₂ =distance travelled in 2nd second
S₃ =distance travelled in 3rd second
S₁ =distance travelled in 1st second
S₂ =distance travelled in 2nd second
S₃ =distance travelled in 3rd second
.
.
.
Sₙ = distance travelled in nth second
Using 2nd equation of motion, the distance travelled in a specific second for a free-falling object is given by (vi = 0 m/s starting of motion from rest potion)
Sₙ = 12g (t2n - t2n-1)
S₁ = 12g t₁2 {for n=1}
S₂ = 12g (t₂2 - t21) {for n=2}
S₃ = 12g (t₃2 - t22) {for n=3}
Now the required ratio is
S₁ : S₂ : S₃ = (1): (4-1): (9-4) { dividing each ratio by 12 g}
or
S₁ : S₂ : S₃ = 1: 3: 5 ------------------Ans
************************************
************************************
Shortcut Links For
1. Website for School and College Level Physics 2. Website for School and College Level Mathematics 3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write to me in the comment box below for any query and also Share this information with your class-fellows and friends.
1. Website for School and College Level Physics
2. Website for School and College Level Mathematics
3. Website for Single National Curriculum Pakistan - All Subjects Notes
© 2022-Onwards by Academic Skills and Knowledge (ASK)
Note: Write to me in the comment box below for any query and also Share this information with your class-fellows and friends.
0 Comments
If you have any QUESTIONs or DOUBTS, Please! let me know in the comments box or by WhatsApp 03339719149