Question

The position of a particle as a function of time t is given by
x=5t³-12t
then acceleration of the particle
Options
1. Remains constant
2. Increases with time
3. Decreases with time
4. First increases and then decreases with time
please answer fast which option is correct ​

Answer 1

Answer:

124566890

Explanation:

1234567890

Answer 2

⚫⚪Answer:⚪⚫

⚪Given:⚪

• $x = 5t^3 -12t$

⚫Solution:⚫

$x = 5t^3 - 12 t$

$\dfrac{dx}{dt} = \dfrac{d}{dt}5t^3 - 12t$

$\boxed{\dfrac{d}{dt}ax^n = nax^{n-1}}$

$\therefore v = 15t^2 - 12$

$\dfrac{dv}{dt} = \dfrac{d}{dt}15t^2 - 12$

$\boxed{\dfrac{dv}{dt} = a}$

$\therefore a = 30t - 12$

$Thus , \ a \ is\ proportional \ to \ 30t$

$Thus \ it \ increases \ with \ time$

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