The following relation [in MKS units] gives the distance traversed s by an accelerated particle of mass $M : s = 6 t + 3 t^{2}$ . The particle's velocity after $2$ seconds in the appropriate unit is

$6$

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$12$

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$18$

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$24$

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Solution

The correct option is C
$18$

Step: 1 Given data:
The given equation of motion is $s = 6 t + 3 t^{2}$ .
where $s =$ displacement, and $t =$ time.
Step: 2 Formula used:
The velocity of the particle $(v) = \frac{ds}{dt}$
Step: 3 Calculate velocity:
$v = \frac{ds}{dt} = \frac{d (6 t + 3 t^{2})}{dt} = 6 + 6 t$ .
Velocity after $2$ sec
$\begin{array}{rcl} v_{t = 2} & = & 6 + 6 (2) \\ = & 18 m / s \end{array}$
Hence option C is correct.

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The following relation [in MKS units] gives the distance traversed s by an accelerated particle of mass M:s=6t+3t2. The particle's velocity after 2 seconds in the appropriate unit is

The correct option is C 18Step: 1 Given data:The given equation of motion is s=6t+3t2.where s=displacement, and t=time.Step: 2 Formula used:The velocity of the particle v=dsdtStep: 3 Calculate velocity:v=dsdt=d6t+3t2dt=6+6t.Velocity after 2secvt=2=6+62=18m/sHence option C is correct.

The following relation [in MKS units] gives the distance traversed s by an accelerated particle of mass $M : s = 6 t + 3 t^{2}$ . The particle's velocity after $2$ seconds in the appropriate unit is