Question

A particle is moving in a straight line along the axis its position is given by x =2t2+2t+4 where x is in metre and t is in second the acceleration of particle is?

Answer 1

Answer:

4 m / sec²

Explanation:

x = 2 t² + 2 t + 4     where x is in meter and t is in second .

We know velocity is rate of o displacement with unit time .

v = d x / d t

Now finding velocity of particle :

v = d ( 2 t² + 2 t + 4 ) / d t

v = 4 t + 2

We also know acceleration is rate change of velocity w.r.t. time .

a = d v / d t

a = d ( 4 t + 2 ) d t

a = 4 m / sec²

Thus the acceleration of particle is 4 m / sec² .

Answer 2

Solution:

We know that,

$\sf{\implies Velocity=\dfrac{dx}{dt}}$

$\sf{\implies \dfrac{d}{dt}(2t^{2}+2t+4)}$

$\sf{\implies \dfrac{d}{dt}(2t^{2})+\dfrac{d}{dt}(2t)+\dfrac{d}{dt}(4)}$

$\sf{By\;using\;identity,\;\dfrac{d}{dx}x^{n}=nx^{n-1}}$

Now,

$\sf{\implies \dfrac{d}{dt}(2t^{2})+\dfrac{d}{dt}(2t)+\dfrac{d}{dt}(4)}$

$\sf{\implies 2(2t^{2-1})+2(t^{1-1})+0}$

${\boxed{\boxed{\bf{\implies velocity = 4t+2}}}}$

Now, we calculate acceleration,

We know that,

$\sf{Acceleration=\dfrac{d(v)}{dt}}$

$\sf{\implies \dfrac{d}{dt}(4t+2)}$

$\sf{\implies \dfrac{d}{dt}(4t)+ \dfrac{d}{dt}(2)}$

$\sf{By\;using\;identity,\;\dfrac{d}{dx}x^{n}=nx^{n-1}}$

Now,

$\sf{\implies \dfrac{d}{dt}(4t)+ \dfrac{d}{dt}(2)}$

$\sf{\implies 4(t^{1-1})+0}$

${\boxed{\boxed{\bf{\implies acceleration = 4\;m/s^{2}}}}}$