x coordinate and y coordinate of a particle is x=(2t^3)/3m, y= (t^2+t)

1.	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» 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. $\frac{dy}{dt} = \frac{d}{dt} (t²+t) = 2t + 1$ $\frac{d²y}{dt²} = \frac{d}{dt} (2t+1) = 2$ $\frac{dx}{dt} = \frac{2}{3} \frac{d}{dt}(t³) = 2t²$ $\frac{d²x}{dt²} = 2 \frac{d}{dt}(t²) = 4t$ $⃗a = \frac{d²x}{dt²}i + \frac{d²y}{dt²}j$ $⃗a = 4t i + 2j$ $⃗a(1) = 4*1 i + 2 j$ $⃗a(1) = 4i + 2j$ $\| ⃗a(1)\| = √(4² + 2²)$ $\| ⃗a(1)\| = √20 m/s²$