At time t- 0- the position vector of a partiele moving in the x-y plane is 5 - m- By time t - 0-02 s- its position vector has become 5-1-0-4- m determine the magnitude vav of the average velocity during this interval and the angle made by the average velocity with the positive x-axis-

r⃗ ⃗_⃗ ⃗1⃗ =5î r⃗ ⃗_⃗ ⃗2⃗ =5.1î ̂ ̂ ̂#̂1̂0̂;̂ +0.4ĵ v⃗ ⃗_⃗ ⃗a⃗v⃗ =r⃗ ⃗_⃗ ⃗2⃗ #10; r⃗ ⃗_⃗ ⃗1⃗ #160; / t = 5.1î +0.4ĵ #10; 5î #160; / 0.02 ...

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Byju's Answer

Standard XII

Physics

1st Equation of Motion

At time t= 0,...

Question

At time t= 0, the position vector of a partiele moving in the x-y plane is $5^i$ m. By time t = 0.02 s, its position vector has become $(5.1^i + 0.4^j)$ m determine the magnitude $v_{a v}$ of the average velocity during this interval and the angle made by the average velocity with the positive x-axis.

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Solution

$_{1} = 5^i_{2} = 5.1^i + 0.4^j_{a v} = \frac{_{2} -_{1}}{t} = \frac{5.1^i + 0.4^j - 5^i}{0.02} = 5^i + 20^j |_{a v} | = \sqrt{5^{2} + 20^{2}} = 20.6 m / s t a n θ = \frac{20}{5} = 4 \Rightarrow θ = {t a n}^{- 1} 4$

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At time t= 0, the position vector of a partiele moving in the x-y plane is 5^i m. By time t = 0.02 s, its position vector has become (5.1^i+0.4^j) m determine the magnitude vav of the average velocity during this interval and the angle made by the average velocity with the positive x-axis.

→r1=5^i→r2=5.1^i+0.4^j→vav=→r2−→r1t=5.1^i+0.4^j−5^i0.02=5^i+20^j|→vav|=√52+202=20.6m/stanθ=205=4⇒θ=tan−14

$_{1} = 5^i_{2} = 5.1^i + 0.4^j_{a v} = \frac{_{2} -_{1}}{t} = \frac{5.1^i + 0.4^j - 5^i}{0.02} = 5^i + 20^j |_{a v} | = \sqrt{5^{2} + 20^{2}} = 20.6 m / s t a n θ = \frac{20}{5} = 4 \Rightarrow θ = {t a n}^{- 1} 4$