Q.The displacement of a particle at time $t$ is given by $s=2t^3-5t^2+4t-3$. Find the velocity and displacement at the time when the acceleration is $14\ \text{ft/sec}^2$.

Many quantities in geometry and physics are functions of time (or of some other varying quantity), and the derivative measures how fast one quantity changes as another changes. If $y=f(x)$ and both $x$ and $y$ vary with time $t$, then $dy/dt=(dy/dx)\cdot(dx/dt)$ by the chain rule — this lets us relate the rate of change of $y$ to the rate of change of $x$ once we know how $y$ depends on $x$. Typical situations combine a geometric formula (area of a circle, volume of a sphere or cone, area of a triangle, etc.) with a given rate for one variable to find the rate of another variable at a specific instant. For problems involving similar triangles (a lamp-post and a walking person's shadow, a ladder sliding down a wall), first set up an algebraic relation between the two changing lengths using similar triangles or the Pythagorean theorem, then differentiate that relation wi …