Question

the position x of a particle varies with t as x=at²–bt³. calculate the acceleration after 2 seconds? ​

Answer 1

Position of the particle varies with time $\sf{t}$ as,

$\sf{\longrightarrow x=at^2-bt^3}$

Then velocity of the particle is,

$\sf{\longrightarrow v=\dfrac{dx}{dt}}$

$\sf{\longrightarrow v=\dfrac{d}{dt}\left(at^2-bt^3\right)}$

$\sf{\longrightarrow v=2at-3bt^2}$

Then acceleration of the particle is,

$\sf{\longrightarrow \alpha=\dfrac{dv}{dt}}$

$\sf{\longrightarrow \alpha=\dfrac{d}{dt}\left(2at-3bt^2\right)}$

$\sf{\longrightarrow \alpha=2a-6bt}$

After 2 seconds,

$\sf{\longrightarrow \alpha=2a-6b(2)}$

$\sf{\longrightarrow\underline{\underline{\alpha=2a-12b}}}$

Answer 2

Question:-

The position x of a particle varies with t as x = at² - bt³. Calculate the acceleration after 2 seconds ?

Answer:-

Here, Instantaneous acceleration is denoted by α

And value of instantaneous acceleration at (t = 2) is denoted by α₂

x = at² - bt³

Velocity = dx/dt

→ v = d(at² - bt³)/dt

→ v = [a * 2 * t^(2 - 1)] - [b * 3 * t^(3 - 1)]

→ v = 2at - 3bt²

Acceleration = dv/dt

→ α = d(2at - 3bt²)/dt

→ α = [2 * a * 1 * t^(1 - 1)] - [3 * b * 2 * t^(2 - 1)]

→ α = 2a - 6bt

Value of α at (t = 2) :-

α₂ = 2a - 6b(2)

→ α₂ = 2a - 12b

Ans. 2a - 12b