Question

the acceleration of a particle in m/s2 is given by a=3t2+2t+2 , where t is in sec find the velocity of the particle at the end of 3 second​

Answer 1

Answer:

a= 3t2+2t+2

dv=adt

v=33t3+22t2+2t

v=t3+t2+2t

at t=0,u=2

att=2

v=t3+t2+2t+2

=8+4+2×2+2

=8+4+4+2

=18m/s {answer}

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Answer 2

Given :

Acceleration of a particle in m/s² is given by a=3t²+2t+2 , where t is in sec

To Find :

Velocity of the particle at the end of 3 seconds if particle starts with velocity v = 2 m/s at t = 0 s

Solution :

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★ Acceleration : It is defined as rate of change in velocity

$\sf a=\dfrac{\Delta v}{\Delta t}$

At particular instant ,

$\sf a=\tiny{\Delta t\to 0}\ \footnotesize {\dfrac{\Delta v}{\Delta t}}$

$\implies \sf a=\dfrac{dv}{dt}$

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$\sf a=\dfrac{dv}{dt}$

$\to \sf dv=a\ dt$

Integrating on both sides ,

$\to \sf \int dv=\int a\ dt$

$\to \sf v=\int (3t^2+2t+2)\ dt$

$\to \sf v=t^3+t^2+2t+c$

At t = 0 s , v = 2 m/s

⇒ v = c = 2 m/s

At t = 3 s ,

$\to \sf v=(3)^3+(3)^2+2(3)+(2)$

$\to \sf v =27+9+6+2$

$\to \textsf{\textbf {\pink{v\ =\ 44\ m/s}}}\ \; \bigstar$