Inverse/implicit relationship between position and time
A question where time is given as a function of distance/position (or position is given implicitly), requiring implicit differentiation or algebraic manipulation to find velocity, acceleration, or their dependence on position.
A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant $= \mathrm { k } > 0$ ). Find the time after which the cone is empty.
The relation between time $t$ and distance x is $\mathrm{t} = a\mathrm{x}^2 + \mathrm{bx}$ where a and b are constants. The acceleration is (1) $-2abv^2$ (2) $2bv^3$ (3) $-2av^3$ (4) $2av^2$
A particle located at $x = 0$ at time $t = 0$, starts moving along the positive $x$-direction with a velocity '$v$' that varies as $v = \alpha \sqrt{x}$. The displacement of the particle varies with time as (1) $t^{3}$ (2) $t^{2}$ (3) $t$ (4) $t^{1/2}$
The distance $x$ covered by a particle in one dimensional motion varies with time $t$ as $x ^ { 2 } = a t ^ { 2 } + 2 b t + c$. If the acceleration of the particle depends on $x$ as $x ^ { - n }$, where $n$ is an integer, the value of $n$ is $\_\_\_\_$