No products in the cart.

A car starting from rest accelerates …

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

A car starting from rest accelerates at the rate of 'f' through a distance,  then continue at constant speed for sometime 't' and then decelerates at rate 'f÷2' to come to rest . If total distance is '5s' then prove that s=ft^2÷2.

  • 1 answers

Sunil C N 8 years, 4 months ago

v2-u2=2as (use this formula)

{tex}s = {v^2 \over 2f}{/tex} 

s2=v*t

s3={tex} {v^2 \over f}{/tex}  ,s3=2*s 

total distansce=5s

s+{tex}t* {\sqrt{s*2f}}{/tex}+2s=5s

{tex}t* {\sqrt{s*2f}}=2s{/tex}

{tex}s=ft^2/2{/tex}

 

https://examin8.com Test

Related Questions

Project report
  • 0 answers
2d+2d =
  • 1 answers
√kq,qpower2 R2
  • 0 answers
1dyne convert to S.I unit
  • 1 answers
Ch 1 question no. 14
  • 0 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App