Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 11
`        A body starting from rest moves with constant acceleration.the ratio of distance covered by the body during the 5th sec to that covered in 5 sec `
10 months ago

## Answers : (2)

chetan jangir
101 Points
```							IT is very simple to deal with such problems. We start with basics we can write distance travelled in 5th second is equal to total covered distance in 5 seconds minus total covered distance in 4 seconds.we get $1/2a5^2 -1/2 a4^2$( we have let acceleration  a  it will also be cancelled out in taking ratio) We get 4.5a .And as we know distance covered in first 5 seconds is 1/2a$5^2$ that is 12.5a. Taking ratio we get 9/25.Which is required  answer .Note .We can generalise a formula for distance covered in nth second it comes u/2 +a/2(2n-1)  where u=initial  velocity a = constant acceleration.
```
10 months ago
Saurabh Koranglekar
askIITians Faculty
6115 Points
```							Dear studentS5 = 0.5a(25)S5th = 0.5a( 25-16) = 0.5a(9)Ratio s5th/ s5 = 9 /25Regards
```
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

## Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details