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3-2.Motion in Plane
medium
The position vector of a particle is determined by the expression $\vec r = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$ The distance traversed in first $10 \,sec$ is........ $m$
A$500$
B$300$
C$150$
D$100$
Solution
(a) $\vec r = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$
at $t = 0$, ${\vec r_1} = 7\hat k$
at $t = 10\sec $, ${\vec r_2} = 300\hat i + 400\hat j + 7\hat k$,
$\overrightarrow {\Delta r} = {\vec r_2} – {\vec r_1} = 300\hat i + 400\hat j$
$|\overrightarrow {\Delta r} |\, = \,|{\vec r_2} – {\vec r_1}|\, = \sqrt {{{(300)}^2} + {{(400)}^2}} = 500m$
at $t = 0$, ${\vec r_1} = 7\hat k$
at $t = 10\sec $, ${\vec r_2} = 300\hat i + 400\hat j + 7\hat k$,
$\overrightarrow {\Delta r} = {\vec r_2} – {\vec r_1} = 300\hat i + 400\hat j$
$|\overrightarrow {\Delta r} |\, = \,|{\vec r_2} – {\vec r_1}|\, = \sqrt {{{(300)}^2} + {{(400)}^2}} = 500m$
Standard 11
Physics
