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4-1.Newton's Laws of Motion
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The position vector of a particle related to time $t$ is given by $\overrightarrow{ r }=\left(10 t \hat{ i }+15 t ^2 \hat{ j }+7 \hat{ k }\right) m$ The direction of net force experienced by the particle is
A
Positive $y$-axis
B
Positive $x$-axis
C
Positive $z-$axis
D
In $x-y$ plane
(JEE MAIN-2023)
Solution
$\overrightarrow{ r }=10 t \hat{ i }+15 t ^2 \hat{ j }+7 \hat{ k }$
$\overrightarrow{ v }=10 \hat{ i }+30 t \hat{ j }$
$\overrightarrow{ a }=30 \hat{ j }$
So Net force is along $+ y$ direction
Standard 11
Physics
