Question

Position of a particle is given as x = 2t3 - 12t2 + 2,
find the acceleration when particle is at rest​

Answer 1

Answer:

As per given

$x = 2 {t}^{3} - 12 {t}^{2} + 2 \\ so \\ \frac{dx}{dt} = v = 6 {t}^{2} - 24t \\ if \: body \: is \: at \: rest \: then \\ v = 0 \: \: and \\ 6 {t}^{2} - 24t = 0 \\ 6t(t - 4) = 0 \\ so \\ t = 0 \: and \: 4 \\ \\ also \\ \frac{dv}{dt} = a = 12t - 24 \\ a = 12(4) - 24 \\ a = 48 - 24 = 24m {s}^{ - 2}$

Answer 2

Answer:

$position \: of \: particle = 2t {}^{3 } - 12t {}^{2} + 2$

Now for finding acceleration from this relation we should find the second derivative of this because in first derivative we get the value of velocity so double differentiate it.

After double differentiation we get

acceleration= 12t-24

Now the particle is at rest so t=0

therefore, acceleration at rest = -24m/s^2

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