Please scroll down to get to the study materials.
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.
\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*}
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:
$$\vec V_AB = \vec V_A - \vec V_B $$
$$\vec V_AB = \vec V_A - (- \vec V_) = \vec A + \vec B$$
\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.
ASK ANY QUESTION ON Equation of Motion with Uniform Acceleration and Relative Velocity
You must login to reply
aakriti
relative velocity in two dimensioms
Jan 28, 2017
1 Replies
Successfully Posted ...
Please Wait...