Question

x=2t3-4t2+3t+4
find the Initial velocity and initial acceleration​

Answer 1

$\huge{\mathcal{\underline{\green{SOLUTION }}}}$

$x \: = 2 {t}^{3} - 4 { t}^{2} + 3t - 4$

$\implies \: \frac{dx}{dt} = v \: = 6 {t}^{2} - 8t + 3$

$\implies \: \frac{ {d}^{2}x }{d {t}^{2}} = 12t - 8$

So

Initial velocity

$6 \times {0}^{2} - 8 \times 0 + 3 = 3 \: \: unit$

Initial acceleration

$= 12 \times 0 - 8 = - 8 \: unit$

Answer 2

Answer:

x=2t3−4t2+3t−4

\implies \: \frac{dx}{dt} = v \: = 6 {t}^{2} - 8t + 3⟹dtdx=v=6t2−8t+3

\implies \: \frac{ {d}^{2}x }{d {t}^{2}} = 12t - 8⟹dt2d2x=12t−8

So

Initial velocity

6 \times {0}^{2} - 8 \times 0 + 3 = 3 \: \: unit6×02−8×0+3=3unit

Initial acceleration

= 12 \times 0 - 8 = - 8 \: unit=12×0−8=−8unit