Physics Two Dimensional Motion Question 142
Question: The coordinates of a particle moving in x-y plane at any instant of time t are $ \text{x = 4}{{\text{t}}^{\text{2}}}\text{; y = 3}{{\text{t}}^{\text{2}}} $ . The speed of the particle at that instant is
Options:
A) 10 t B)5 t C) 3 t D)2 t
Show Answer
Answer:
Correct Answer: A
Solution:
[a] According to the question, at any instant t, $ x=4t^{2},y=3t^{2} $ $ \
Therefore {v _{x}}=\frac{dx}{dt}=\frac{d}{dt}( 4t^{2} )=8t $ $ \text{and }{v _{y}}=\frac{dy}{dt}=\frac{d}{dt}( 3t^{2} )=6t $
$ \text{and }{{v _{y}}=\frac{dy}{dt}=\frac{d}{dt}( 3t^{2} )=6t $ The speed of the particle at instant t.. $ \text{v},\text{=},\sqrt{v _{x}^2 +v _{y^2}},\text{=},\sqrt{{{( \text{8t} )}^{\text{2}}}\text{+}{{( \text{6t} )}^{\text{2}}}}\text{=10},\text{t} $
BETA
