A car, starting from rest, accelerates at the rate $f$ through a distance $S$, then continues at constant speed for time $t$ and then decelerates as the rate $f / 2$ to come to rest. If the total distance travelled is $15 S$, then
The velocity-time graph for the given situation can be drawn as below. Magnitudes of slope of $O A=f$
and slope of $B C=\frac{f}{2}$$v=f t_{1}=\frac{f}{2} t_{2}$$\therefore t_{2}=2 t_{1}$
In graph area of $\triangle O A D$ gives
distances, $ S=\frac{1}{2} f t_{1}^{2}$ ... (i)
Area of rectangle $A B E D$ gives distance travelled in time $t$.
$S_{2}=\left(f t_{1}\right) t$
Distance travelled in time $t_{2}$$=S_{3}=\frac{1}{2} f_{2}\left(2 t_{1}\right)^{2}$
Thus, $ S_{1}+S_{2}+S_{3}=15 S$$S+\left(f t_{1}\right) t+f t_{1}^{2}=15 S$$S+\left(f t_{1}\right) t+2 S =15 S \left(S=\frac{1}{2} f t_{1}^{2}\right) $$\left(f t_{1}\right) t =12 S $ ... (ii)
From Eqs. (i) and (ii), we have
$\frac{12 S}{S}=\frac{\left(f t_{1}\right) t}{\frac{1}{2}\left(f t_{1}\right) t_{1}} $$t_{1}=\frac{t}{6}$
From E (i), we get
$\therefore S=\frac{1}{2} f\left(t_{1}\right)^{2} $$S=\frac{1}{2} f\left(\frac{t}{6}\right)^{2}=\frac{1}{72} f t^{2}$
The motion in a straight lineis an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
Types of Linear Motion:
Linear motion is also known as the Rectilinear Motion which are of two types:
Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.
