Subject: Physics

Suppose a body moving in a straight line. Let u be its initial velocity and in time t, it increases to v moving with a uniform acceleration a between these two points.

The accelerated of the body is

\begin{align*} a &= \frac {v-u}{t} \\ \text {or,} at &= v-u \dots (i) \\ \text {The average velocity is} \\ v_{av} &= \frac {u+v}{2} \dots (ii) \\ \text {If s is the displacement, then} \\ s &= v_{av} \times t \\ &= \frac {u+v}{2} \times t \\ &= \frac {u+u+at}{2} \times t \\ &= \frac {2ut + at^2}{2} = ut + \frac 12 at^2 \\ \therefore s &= ut + \frac 12 at^2 \dots (iii) \\ \text {Squaring equation} (i), \text {we get} \\ v^2 &= (u+at)^2 \\ &= u^2 +2aut + a^2t^2 \\ &= u^2 + 2a(ut + \frac 12 at^2) \\ \therefore v^2 &= u^2 + 2as \dots (iv) \end{align*}

The above equations of motion can be derived from the velocity-time graph as shown in the figure.

Consider an object moving with uniform acceleration (a) on a straight line. Let u be the initial velocity at t=0 and v be the final velocity after time t. As shown in figure OA = ED = u; OC = EB = v, OE = t= AD.

- $$V = u + at$$

We know that slope of velocity-time graph of uniformly accelerated motion represents the acceleration of the object.

\begin{align*} \therefore \text {acceleration} &= \text {slope of the velocity-time graph AB} \\ \text {or,} \: a &= \frac {DB}{AD} = \frac {DB}{OE} =\frac {EB – ED}{OE} = \frac {v – u}{t} \\ \text {or,} v – u &= at \\ \text {or,} v &= u + at \\ \end{align*}

- $$S = ut + \frac 12 at^2$$

We know that the area under the velocity-time graph for a given time interval represents the distance covered by a uniformly accelerated object in a given time interval.

\begin{align*} \text {From graph, acceleration, a} &= \text {slope of velocity time graph AB.} \\ \therefore a &= \frac {DB}{AD} = \frac {DB}{t} \\ \text {or,} DB &= at \\ \text {Now distance travelled by object in time t,} \\ s &= \text {area of trapezium OABE} = \text {area of rectangle OADE} + \text {area of triangle ADB} \\ &= OA \times OE + \frac 12 OB \times AD \\ &= ut + \frac 12 at \times t = ut + \frac 12 at^2 \\ \text {therefore,} s &= ut + \frac 12 at^2. \end{align*} - $$v^2 = u^2 + 2as $$

\begin{align*} \text {Distance travelled by the object in time interval t is} \\ s &= \text {area of trapezium OABF} \\ &= \frac 12 (EA + OA) \times OE = \frac 12 (EA + ED) \times OE \dots {i} \\ \text {Acceleration} &= \text {slope of velocity time graph AB} \\ \text {or,} a &= \frac {DB}{AD} = \frac {EB – ED}{OE} \\ \text {or,} OE &= \frac {EB – ED}{a} \\ \text {putting this value in equation} (i), \text {we get} \\ s &= \frac 12 (EB + ED) \times \frac {EB – ED}{a} \\ &= \frac {1}{2a} (EB^2 –ED^2) \\ &= \frac {1}{2a} (v^2 –u^2) \\ \therefore v^2 = u^2 + 2as \end{align*}

If the acceleration of the body is uniform, distance travelled in t second is

\begin{align*} s_1 &= ut + \frac 12 at^2 \\ \text {distance travelled in (t-1) second is} \\ s_{t-1} &= u(t-1) + \frac 12 a(t-1)^2 \\ \text {Then, distance travelled in between t and (t-1) second is} \\ s_t –s_{t-1} &= \left [ ut + \frac 12 at^2 \right ] - \left [ u(t-1) +\frac 12 a(t-1)^2 \right ] \\ &= ut + \frac 12 at^2 –ut + u - \frac 12 at^2 + at - \frac 12 a \\ &= u + at - \frac a2 \\ \therefore \text {Distance travelled in} t^{th} \text {second is,} \\ s_{th} &= u + a\left ( t - \frac 12 \right ) = u + \frac a2 (2t -1) \end{align*}

**Motion of Freely Falling Bodies**

If air resistance is neglected and a body is dropped from a small height, h above the earth’s surface, it falls with constant acceleration g called the acceleration due to gravity. So the equation of motion are written as

\begin{align*} v &= u + gt \\ h &= ut + \frac 12 at^2 \\ v^2 &= u^2 + 2gh \end{align*}

If the motion is vertically upward, the acceleration of the body is directed upward and is equal to –g. Then equations of motion become

\begin{align*} v &= u - gt \\ h &= ut - \frac 12 gt^2 \\ v^2 &= u^2 - 2gh \end{align*}

The relative velocity of one with respect to the other is to bring the observer at rest and to impose the velocity of observer to the velocity of the object being observed. So, relative velocity is the time rate of change of position of one object with respect to the another object. The basic rule to determine the relative velocity are as follow:

- When two bodies A and B are moving in the same direction with velocity \(\vec A\) and \(\vec B\) respectively, then relative velocity of A with respective to B is

$$\vec V_AB = \vec V_A - \vec V_B $$

- When they are moving in opposite direction, then relative velocity of A with respect to B is

$$\vec V_AB = \vec V_A - (- \vec V_) = \vec A + \vec B$$

- When two bodies are moving making an acute angle θ (90>θ>0), to obtain the relative velocity of an object with respect to an observer, the direction of the observer is reserved and parallelogram is completed. The diagonal of the parallelogram is completed. The diagonal of the parallelogram gives the relative velocity. Similarly, triangle law of vectors can be used to obtain the relative velocity. For example, let two bodies A and B are moving with velocities v
_{A}and v_{B}in two different directions making an angle (θ) as shown in the figure. To find the velocity of B with respect A, the velocity of A is reserved as shown in the figure. Then taking \(\vec v_b\) and -\(\vec V_A\) as two sides of a parallelogram, the diagonal gives the relative velocity as shown in the figure. Similarly the relative of A with respect to B is obtained by reversing the direction of B and the diagonal of the parallelogram gives the relative velocity v_{AB}as shown in the figure. **Raining falling vertically downward**

When a man is stationary and rain is falling vertically downward, the man has to held the umbrella vertically to save himself from the rain. The relative velocity of rain with respect to the man is in the vertical direction.

Suppose rain is falling vertically downward with velocity v and man is walking in the horizontal direction with velocity u. To find the relative velocity of rain with respect to the man, take OQ = v and OM = u. Here ON represents an equal and opposite velocity of the man. The resultant R is represented in magnitude and direction by the vector OS as shown in the figure. Suppose R makes an angle () with the vertical. From the figure, we have

\begin{align*} \tan \theta &= \frac uV \\ \text {or,}\: \theta &= \tan ^{-1} \frac uv \text {Thus, a man walking in rain should hold his umbrella making an angle} \theta \text {with the vertical such that} \theta &= \tan ^{-1} \frac uv \end{align*}

This also explains that why the front wind screen of an automobile moving in rain gets wet where as the hind screen remains dry.

S=ut+12at

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

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