Question

if x=t³+3t²-2t+2 find distance velocity and acceleration in 1 second​

Answer 1

To FinD :-

• The velocity and acceleration at 1s.

AnsweR :-

Here Displacement is given by ,

$\green{\sf:\implies x = t^3+3t^2-2t + 2 }$

And we need to find the velocity and acceleration at t = 1 second. So for that differenciate both sides with respect to t . First order of differenciation will give velocity and the second order of differenciation will give acceleration .

$\purple{\bigstar}\underline{\boldsymbol{ Differenciating \ both \ sides \ wrt \ t :- }}$

• First order of differenciation will give Velocity .

$\sf:\implies \pink{ x = t^3+3t^2-2t + 2} \\\\\sf:\implies \dfrac{dx}{dt}=\dfrac{ d( t^3+3t^2-2t + 2)}{dt} \\\\\sf:\implies \dfrac{dx}{dt}= 3t^{(3-1)}+3.2t^{(2-1)} - 2t + 0 \\\\\sf:\implies v = 3t^2+6t -2 \\\\\sf:\implies v_{(At \ t = \ 2 )}= 3(1)^2 + 6(1)-2\\\\\sf:\implies v_{(At \ t = \ 2 )}= 3 + 6 -2 \\\\\sf:\implies\underset{\blue{\sf Required \ Velocity }}{\underbrace{\boxed{\pink{\frak{ Velocity_{(At \ t = \ 2 )}= 7m/s }}}}}$

$\rule{200}2$

$\purple{\bigstar}\underline{\boldsymbol{ Again \ Differenciating \ both \ sides \ wrt \ t :- }}$

• Second order of differenciation will give acclⁿ.

$\sf:\implies\pink{ v = 3t^2+6t -2}\\\\\sf:\implies \dfrac{dv}{dt}= \dfrac{ d( 3t^2+6t-2)}{dt}\\\\\sf:\implies \dfrac{dv}{dt}= 2.3t^{(2-1)} +6(1) + 0 \\\\\sf:\implies a = 6t + 6 \\\\\sf:\implies a_{(At \ t = \ 1 )}= 6(1)+6 \\\\\sf:\implies a_{(At \ t = \ 1 )}= 6 + 6 \\\\\sf:\implies\underset{\blue{\sf Required \ Acceleration}}{\underbrace{\boxed{\pink{\frak{ Acceleration_{(At \ t = \ 1 )}= 12m/s^2}}}}}$