669 Views

a particle is moving such that its position coordinates (x, y)are (2m, 3m)at time t=0, (6m, 7m)at time t=2s and (13m, 14m)at time t=5s.average velocity vector (v)from t=0to t=5s is.


Varnika 30th Apr, 2020
Answer (1)
ADITYA KUMAR Student Expert 30th Apr, 2020
Hi

We know that average velocity is given by

= net displacement total time

Now,
net displacement = final position - intial position
= (13i + 14j) - (2i +3j)
= (13i - 2i) + (14j - 3j)
= 11i + 11j

average velocity = (11 i + 11j)/5

Then the magnitude of average velocity
= (121/25+ 121/25 ) ^1/2

= 11(2/5 )^1/2

Thankyou
1 Comment
Comments (1)
30th Apr, 2020
Thanks
Reply

Related Questions

Amrita University B.Tech 2026
Apply
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
UPES B.Tech Admissions 2026
Apply
Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
UPES Integrated LLB Admission...
Apply
Ranked #18 amongst Institutions in India by NIRF | Ranked #1 in India for Academic Reputation by QS Rankings | 16 LPA Highest CTC
Great Lakes Institute of Mana...
Apply
Admissions Open | Globally Recognized by AACSB (US) & AMBA (UK) | 17.8 LPA Avg. CTC for PGPM 2025
Jain University, Bangalore - ...
Apply
NAAC A++ Approved | Curriculum Aligned with BCI & UGC
Amity University, Mumbai Law ...
Apply
Ranked as India’s #1 Not for profit pvt. University by India Today
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