A particle starts from rest and moves with an acceleration of -4-t-3- ms2 where -t- is in seconds- The velocity of the particle when it is 2 m from the origin is

According to given information, a=4 (y 3) t ≤ 3 s =4+(t 3) t gt;3 s From a=d vd t nbsp;⇒∫ d v=∫ a d t nbsp;⇒∫_0^v d v=∫_0^3(7 t) d t+∫(1+t) d t nbsp;⇒ v=21 m ...

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Byju's Answer

Standard IX

Physics

The Equations of Motion

A particle st...

Question

A particle starts from rest and moves with an acceleration of $= [4 - | t - 3 |] \frac{m}{s^{2}}$ where $^{'} t^{'}$ is in seconds. The velocity of the particle when it is $2$ m from the origin is

$16.5 m / s$

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$21 / m / s$

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$12.5 m / s$

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$4.5 m / s$

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Solution

The correct option is B
$21 / m / s$

According to given information,
$a = 4 - (y - 3) t \leq 3 s$
$= 4 + (t - 3) t > 3 s$
$From a = \frac{d v}{d t}$
$\Rightarrow \int d v = \int a d t$
$\Rightarrow \int_{0}^{v} d v = \int_{0}^{3} (7 - t) d t + \int (1 + t) d t$
$\Rightarrow v = 21 m / s$

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A particle starts from rest and moves with an acceleration of =[4−|t−3|]ms2 where ′t′ is in seconds. The velocity of the particle when it is 2 m from the origin is

The correct option is B 21/m/s According to given information, a=4−(y−3)t≤3s =4+(t−3)t>3s From a=dvdt ⇒∫dv=∫adt ⇒∫v0dv=∫30(7−t)dt+∫(1+t)dt ⇒v=21m/s

A particle starts from rest and moves with an acceleration of $= [4 - | t - 3 |] \frac{m}{s^{2}}$ where $^{'} t^{'}$ is in seconds. The velocity of the particle when it is $2$ m from the origin is

The correct option is B
$21 / m / s$

According to given information,
$a = 4 - (y - 3) t \leq 3 s$
$= 4 + (t - 3) t > 3 s$
$From a = \frac{d v}{d t}$
$\Rightarrow \int d v = \int a d t$
$\Rightarrow \int_{0}^{v} d v = \int_{0}^{3} (7 - t) d t + \int (1 + t) d t$
$\Rightarrow v = 21 m / s$