Question

The velocity of a partice is given
by v= 12+3t+21t². find acceleration at t=2second​

Answer 1

Answer:

51 $m$/$s^{2}$

Explanation:

Given,

u= 0$ms^{-2}$ (since the initial velocity of the particle is not given let us assume it     starts from rest)

t= 2 seconds

then,

v= 12+ 3v+ 21v²   {given}

v= 12 + 3×2+ 21×2²

v= 12+ 6+ 84

v= $102ms^{-2}$

∴ Acceleration, a = $\frac{v-u}{2}$

                             = $\frac{102-0}{2}$

                             = $\frac{102}{2}$

                             =$51 ms^{-2}$

Answer 2

Answer:

v= 12 +3t+21t^2  at t=2sec

differentiate with recpect to x

dv\dx=  d\dx(12+3t+21t^2)

a = 0+3+42t

 a= 42(2)+3

 a= 84+3

a= 87