<< Chapter < Page | Chapter >> Page > |
Distance in the interval t = 0 to 1 s is :
$$\begin{array}{l}{s}_{1}=1-0=1\phantom{\rule{2pt}{0ex}}m\\ {s}_{2}=4-0=4\phantom{\rule{2pt}{0ex}}m\end{array}$$
Total distance is 1 + 4 = 5 m.
6: Velocity is zero, when t = 1 s. In this period, displacement is 1 m.
7: In order to determine the nature of force on the particle, we first determine the acceleration as :
$$\begin{array}{l}a=\frac{\u0111v}{\u0111t}=\frac{\u0111}{\u0111t}(2t-2)=2\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$$
Acceleration of the motion is constant, independent of time. Hence, force on the particle is also constant during the motion.
The equation of motions in one dimension for constant acceleration is obtained from the equations of motion established for the general case i.e. for the three dimensional motion. In one dimension, the equation of motion is simplified ( r is replaced by x or y or z with corresponding unit vector). The three basic equations of motions are (say in x - direction) :
$$\begin{array}{l}\mathrm{1:}\phantom{\rule{2pt}{0ex}}\mathbf{v}=\mathbf{u}+\mathbf{a}t\\ \mathrm{2:}\phantom{\rule{2pt}{0ex}}{\mathbf{v}}_{\mathrm{avg}}=\frac{(\mathbf{u}+\mathbf{v})}{2}\\ \mathrm{3:}\phantom{\rule{2pt}{0ex}}\Delta x\mathbf{i}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{t}^{2}\end{array}$$
Significantly, we can treat vector equivalently as scalars with appropriate sign. Following is the construct used for this purpose :
Once, we follow the rules as above, we can treat equations of motion as scalar equations as :
$$\begin{array}{l}\mathrm{1:}\phantom{\rule{2pt}{0ex}}v=u+at\\ \mathrm{2:}\phantom{\rule{2pt}{0ex}}{v}_{\mathrm{avg}}=\frac{(u+v)}{2}\\ \mathrm{3:}\phantom{\rule{2pt}{0ex}}\Delta x=ut+\frac{1}{2}a{t}^{2}\end{array}$$
Problem : A car moving with constant acceleration covers two successive kilometers in 20 s and 30 s respectively. Find the acceleration of the car.
Solution : Let "u" and "a" be the initial velocity and acceleration of the car. Applying third equation of motion for first kilometer, we have :
$$\begin{array}{l}1000=ux20+\frac{1}{2}a{20}^{2}=20u+200a\\ \Rightarrow 100=2u+20a\end{array}$$
At the end of second kilometer, total displacement is 2 kilometer (=2000 m) and total time is 20 + 30 = 50 s. Again applying third equation of motion,
$$\begin{array}{l}2000=ux50+\frac{1}{2}a{50}^{2}=50u+2500a\\ \Rightarrow 200=5u+125a\end{array}$$
Solving two equations,
$$\begin{array}{l}a=-2.86\phantom{\rule{2pt}{0ex}}m/{s}^{2}\end{array}$$
Note that acceleration is negative to the positive direction (direction of velocity) and as such it is termed “deceleration”. This interpretation is valid as we observe that the car covers second kilometer in longer time that for the first kilometer, which means that the car has slowed down.
It is important to emphasize here that mere negative value of acceleration does not mean it to be deceleration. The deciding criterion for deceleration is that acceleration should be opposite to the direction of velocity.
We have observed that when a feather and an iron ball are released from a height, they reach earth surface with different velocity and at different times. These objects are under the action of different forces like gravity, friction, wind and buoyancy force. In case forces other than gravity are absent like in vacuum, the bodies are only acted by the gravitational pull towards earth. In such situation, acceleration due to gravity, denoted by g, is the only acceleration.
Notification Switch
Would you like to follow the 'Physics for k-12' conversation and receive update notifications?