Question

x coordinate and y coordinate of a particle is x=(2t^3)/3m, y= (t^2+t)m find the accleration of the particle at t=1 sec

Answer 1

Answer:

$⃗a(1) = 4i + 2j$

$| ⃗a(1)| = √20 m/s²$

Explanation:

Given

The x co- ordinate of the particle is

$x = \frac{2}{3} {t}^{3} \: m$

The y co-ordinate of the particle is given by

$y = ( {t}^{2} + t) \: m$

Calculation

To calculate acceleration we have to calculate d²y/dt² and d²x/dt².

$\frac{dy}{dt} = \frac{d}{dt} (t²+t) = 2t + 1$

$\frac{d²y}{dt²} = \frac{d}{dt} (2t+1) = 2$

$\frac{dx}{dt} = \frac{2}{3} \frac{d}{dt}(t³) = 2t²$

$\frac{d²x}{dt²} = 2 \frac{d}{dt}(t²) = 4t$

$⃗a = \frac{d²x}{dt²}i + \frac{d²y}{dt²}j$

$⃗a = 4t i + 2j$

$⃗a(1) = 4*1 i + 2 j$

$⃗a(1) = 4i + 2j$

$| ⃗a(1)| = √(4² + 2²)$

$| ⃗a(1)| = √20 m/s²$