NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Physics
- Chapter
- Motion in Two Dimensions
NTA Abhyas 2020 · Physics · Motion in Two Dimensions
Two particles A and B are located at points (0, 10 3 ) and ( 0, 0 )...
Two particles are located at points and ( 0, 0 ) in x - y plane. They start moving simultaneously at the time with constant velocities \overset{\to}{V}A = 5\overset{^}{i}m{s}^{- 1} and \overset{\to}{V}B = - 5\sqrt{3}\overset{^}{i}m{s}^{- 1}, respectively. The time when they are closest to each other is found to be second. Find . All distances are given in meter.
Correct answer
3
Step-by-step solution
As observed by B motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is
{\overset{\to}{V}}_{AB} = 5\overset{^}{i} + 5\sqrt{3}\overset{^}{j}
Time taken \left(t\right) = \dfrac{\left|{\overset{\to}{S}}_{AB}\right|}{\left|{\overset{\to}{V}}_{AB}\right|} = \dfrac{AM}{{V}_{AB}} = \dfrac{10\sqrt{3}cos {30}^{o}}{10}= 1.5s
{\overset{\to}{V}}_{AB} = 5\overset{^}{i} + 5\sqrt{3}\overset{^}{j}
Time taken \left(t\right) = \dfrac{\left|{\overset{\to}{S}}_{AB}\right|}{\left|{\overset{\to}{V}}_{AB}\right|} = \dfrac{AM}{{V}_{AB}} = \dfrac{10\sqrt{3}cos {30}^{o}}{10}= 1.5s