Fewpal
+2 votes
3.3k views
in JEE by (35 points)

1 Answer

+2 votes
by (21.1k points)
selected by
 
Best answer

Solution: We have

Initial velocity, u = 5 m/s due east

Final velocity, v = 5 m/s due north

Change in velocity, Δv = v – u = 5 m/s due north - 5 m/s due east

=> Δv = 5 m/s due north + 5 m/s due west  [negative of east direction is west]

Δv = (v2 + u2)1/2

=> Δv = 5√2 m/s

Direction of Δv is given by,

tanθ = v/u = -1

=> θ = -45o

Thus, the acceleration is a = Δv/t =  5√2/10 = 1/√2 m/s2 and is directed towards North-West direction.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...