A body of mass 2, kg moves under a force of (2 hat{ i }+3 hat{ j }+5 hat{ k }) ,N . It starts

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4-1.Newton's Laws of Motion

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A body of mass $2\, kg$ moves under a force of $(2 \hat{ i }+3 \hat{ j }+5 \hat{ k }) \,N$. It starts from rest and was at the origin initially. After $4 \,s$, its new coordinates are $(8, b, 20)$. The value of $b$ is ........ . (Round off to the Nearest Integer)

A

$16$

B

$8$

C

$12$

D

$20$

(JEE MAIN-2021)

Solution

$\overrightarrow{ a } =\frac{\overrightarrow{ F }}{ m }=\frac{2 \hat{ i }+3 \hat{ j }+5 \hat{ k }}{2}$

$=\hat{ i }+1.5 \hat{ j }+2.5 \hat{ k }$

$\vec{r} =\overrightarrow{ u } t +\frac{1}{2} \overrightarrow{ a } t ^{2}$

$=0+\frac{1}{2}(\hat{ i }+1.5 \hat{ j }+2.5 \hat{ k })(16)$

$=8 \hat{ i }+12 \hat{ j }+20 \hat{ k }$

$b=12$

Standard 11

Physics

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