Calculus 11: Velocity and Acceleration

Let us assume that a particle is moving along a straight line and that the function $s=f(t)$ describes the position of moving particle at the time $t$. In physics, such a function $s=f(t)$ is called a motion.

Suppose the particle passes the points $P$ and $Q$ at the times $t$ and $t+\Delta t$, respectively. If $s$ and $s+\Delta s$ are the respective distances from some fixed point $O$, then the average velocity of the particle during the time interval $\Delta t$ is $$\frac{\Delta s}{\Delta t}=\frac{f(t+\Delta t)-f(t)}{\Delta t}=\frac{\mbox{Distance Traveled}}{\mbox{Time Elapsed}}.$$ The instantaneous velocity $v$ of the particle at the time $t$ is then given by the derivative of motion $s=f(t)$ $$v=\frac{ds}{dt}=\lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}.$$ In physics, the instantaneous velocity is also denoted by $\dot{s}$ or $\dot{f}(t)$. This dot notation was introduced by Sir Issac Newton.

Similarly, if $\Delta v$ is the change in the velocity of the particle as it moves from $P$ to $Q$ during the time interval $\Delta t$, then $$a=\frac{dv}{dt}=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t}$$ is the acceleration of the particle at the time $t$. Using dot notation, the acceleration is also denoted by $\dot{v}$, $\ddot{s}$, $\ddot{f}(t)$, or $\frac{d^2 s}{dt^2}$. The last notation $\frac{d^2 s}{dt^2}$ is due to Gottfried Leibniz.

If a body is thrown vertically upward with a certain initial velocity $v_0$, its distance $s$ from the starting point is given by the formula $$s(t)=v_0t-\frac{1}{2}gt^2,$$ where $g$ is the gravitational constant $g=9.8\mbox{m}/\mbox{sec}^2=32\mbox{ft}/\mbox{sec}^2.$

Example. From the top of a building 96 feet high, a ball is thrown directly upward with a velocity of 80 feet per second. Find

1. the time required to reach the highest point
2. the maximum height attained, and
3. the velocity of the ball when it reaches the ground.
   

Solution. $v_0=80$ ft/sec and $g=32\mbox{ft}/\mbox{sec}^2$, so $$s=80t-16t^2$$ and $$v=\frac{ds}{dt}=80-32t.$$

1. At the heighest point, $v=0$ that is $0=80-32t$. So, $t=\frac{5}{2}$.
2. $s\left(\frac{5}{2}\right)=80\left(\frac{5}{2}\right)-16\left(\frac{5}{2}\right)^2=100$ft. Hence the height of the ball above the ground is 196 feet.
3. Since the ball will reach the ground when $s=-96$, it follows that $-96=80t-16t^2$ or $16(t-6)(t+1)=0$. Hence $t=6$ and the velocity is $v(6)=80-32\cdot 6=-112$ft/sec when the ball strikes the ground. The negative sign merely indicates that the velocity of the ball is directed downward.