A person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$, then the relation between $v , v _1$ and $v _2$ will be :
$v=v_1+v_2$
$v =\frac{ v _1+ v _2}{2}$
$\frac{2}{ v }=\frac{1}{ v _1}+\frac{1}{ v _2}$
$\frac{1}{ v }=\frac{1}{ v _1}+\frac{1}{ v _2}$
Figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of $v$ and $a$ in the three intervals. What are the accelerations at the points $A, B, C$ and $D$?
The posttion of an object moving along $x$ -axis ts given by $x=a+b t^{2}$ where $a=8.5\; \mathrm{m}, b=2.5 \;\mathrm{m} \mathrm{s}^{-2}$ and $t$ is measured in seconds. What is the average velocity between $t=2.0 \;\mathrm{s}$ and $t=4.0 \;\mathrm{s} ?$
The average velocity of a body moving with uniform acceleration travelling a distance of $3.06\, m$ is $ 0.34 ms^{-1}$. If the change in velocity of the body is $ 0.18ms^{-1}$ during this time, its uniform acceleration is.........$ms^{-2}$
The position $(x)$-time $(t)$ graph for a particle moving along a straight line is shown in figure. The average speed of particle in time interval $t=0$ to $t=8 \,s$ is .......... $m / s$
Define speed. Define average speed.