384 Views

An automobile travelling with a speed of  can brake to stop within a distance of 20m. If the car is going twice as fast, i.e., 120 Km/h, the stopping distance will be ________ m


Paul Vijay vardhan 9th Apr, 2020
Answer (1)
Smrity 9th Apr, 2020

Hey,

since the speed is increased twice therefore initial speed is 120/2=60 km/hr.

now, there are two cases:

Case I:

given,

u=60 km/hr

v=0

S=20 m

Applying formula: v^2=u^2+2aS

0=(60*5/18)^2-(2*a*S)

a=(60*5/18)^2/(2*20)

Case 2:

u=120 km/hr

v=0

a=(60*5/18)^2/(2*20) (from case 1)

Applying the same formula as case 1

v^2=u^2+2aS

0=(120*5/18)^2+2*(60*5/18)/(2*20*S)

therefore

S=80 m, which is the required distance and answer.

I hope this helps.

All the best!


Related Questions

Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities in QS Asia Rankings 2025 | Scholarships worth 210 CR
Amity University, Noida Law A...
Apply
700+ Campus placements at top national and global law firms, corporates, and judiciaries
Amity University, Noida BBA A...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
UPES | BBA Admissions 2025
Apply
#41 in NIRF, NAAC ‘A’ Grade | 100% Placement, up to 30% meritorious scholarships | Last Date to Apply: 28th Feb
MAHE Manipal M.Tech 2025
Apply
NAAC A++ Accredited | Accorded institution of Eminence by Govt. of India | NIRF Rank #4
Sanskriti University LLM Admi...
Apply
Best innovation and research-driven university of Uttar Pradesh
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books