Physics - Kinematics Question with Solution | TestHub

Given that

$\frac{dx}{dt}=\frac{dy}{dt}=c$

$\therefore \frac{d^{2}x}{d{t}^2}=\frac{d^{2}y}{d{t}^2}=0$

Further z = ax³+ by²

$\therefore \frac{dz}{dt}=3a{x}^2\frac{dx}{dt}+2by\frac{dy}{dt}$

$=3a{cx}^2+2bcy\left(\frac{dx}{dt}=c=\frac{dy}{dt}\right)$

$\therefore \frac{d^{2}z}{d{t}^2}=6acx\left(\frac{dx}{dt}\right)+2bc\left(\frac{dy}{dt}\right)$

= 6ac²x + 2bc²

Now acceleration of particle is

$\overrightarrow{a}=\frac{d^{2}x}{d{t}^2}\widehat{\text{i}}+\frac{d^{2}y}{d{t}^2}\widehat{\text{j}}+\frac{d^{2}z}{d{t}^2}\widehat{k}$

$=\left(6a{c}^{2}x+2b{c}^2\right)\widehat{k}$