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A particle moves from the point 2i^+4j^ m , at t=0 with an initial velocity 5i^+4j^ m/s. It is acted upon by a constant acceleration 4i^+4j^ m/s2. What is the distance of the particle from the origin at time t=2 s in meters?

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answer is 0028.28.

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Detailed Solution

GivenInitial Position, r→o=2i^+4j^ mInitial velocity, u→=5i^+4j^ m/sAcceleration, a→=4i^+4j^ m/s2S→=u→t+12a →t2⇒r→−r→o=(5i^+4j^)×2+12(4i^+4j^)×4⇒r→=(2i^+4j^)+(10i^+8j^)+8i^+8j^⇒r→=20i^+20j^⇒|r→|=202m⇒|r→|=28.28m

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