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Question: A particle starts from origin at t = 0 with a velocity \(5\widehat{i}ms^{- 1}\)and moves in x-y plan...

A particle starts from origin at t = 0 with a velocity 5i^ms−15\widehat{i}ms^{- 1}and moves in x-y plane under the action of force which produces a constant acceleration of 3i^+2j^ms−23\widehat{i} + 2\widehat{j}ms^{- 2}. The speed of the particle at this time is

A

16ms−116ms^{- 1} (16 m/s)

B

26ms−126ms^{- 1} (26 m/s)

C

36ms−136ms^{- 1} (36 m/s)

D

46ms−146ms^{- 1} (46 m/s)

Answer

26ms−126ms^{- 1} (26 m/s)

Explanation

Solution

Velocity, v→=dr→dt=ddt(5t+1.5t2)i^+1t2j^\overset{\rightarrow}{v} = \frac{d\overset{\rightarrow}{r}}{dt} = \frac{d}{dt}(5t + 1.5t^{2})\widehat{i} + 1t^{2}\widehat{j}

=(5+3t)i^+2tj^= (5 + 3t)\widehat{i} + 2t\widehat{j}

As t=6s,v→=23i^+12j^t = 6s,\overset{\rightarrow}{v} = 23\widehat{i} + 12\widehat{j}

The speed of the particle is

∣v→∣=(23)2+(12)2=26ms−1\left| \overset{\rightarrow}{v} \right| = \sqrt{(23)^{2} + (12)^{2}} = 26ms^{- 1}