Q.The distance - time relationship of a moving body is given by $y = F(t)$ then the acceleration of the body is the :

The derivative $f'(x)$ of a function carries two equivalent readings, and this chapter leans on both throughout.

As a slope. For the curve $y=f(x)$, the slope of the chord joining $(x,f(x))$ and $(x+h,f(x+h))$ is the Newton quotient

Taking $h\to0$ gives the slope of the curve at $(x,f(x))$ itself:

If $\theta$ is the angle the tangent makes with the positive $x$-axis (measured anticlockwise), then $f'(x)=\tan\theta$.

As a rate of change. $f'(x)=\dfrac{dy}{dx}$ is also the instantaneous rate of change of $y$ with respect to $x$; over an interval $[a,b]$ the average rate of change is the ordinary difference quotient $\dfrac{f(b)-f(a)}{b-a}$ (a chord slope), while the derivative at a single point is the instantaneous rate.

Motion along a line. If $s=f(t)$ is the position of an object at time $t$ (measured from a fixed origin, positive direction = forward):

• Speed $=|v(t)|=\left|\dfrac{ds}{dt}\right|$ — always non-negative, regardless of direction.
• $v(t)=0$: the particle is momentarily at rest.
• $v(t)>0$: moving forward; $v(t)<0$: moving backward.
• The particle changes direction exactly where $v(t)$ changes sign (not merely where it is zero — the sign must flip on either side).
• If the particle reverses direction at time $t_c\in(t_1,t_2)$, the total distance travelled from $t_1$ to $t_2$ is $|s(t_c)-s(t_1)|+|s(t_2)-s(t_c)|$ — NOT simply $|s(t_2)-s(t_1)|$, since backtracking would otherwise cancel out.
• Near Earth's surface a freely falling body has constant acceleration $g\approx9.8\ \text{m/s}^2$ ($32\ \text{ft/s}^2$), giving $a=-g,\ v=-gt+v_0,\ s=-\tfrac12gt^2+v_0t+s_0$. …