Science M.C.Q. [1 M]

Let the objects speed be $x$ and $y$ respectively, when moving towards each other:
$x + y = 10 ...(1)$
and, when moving in same direction
$x - y = 6 ...(2)$
now,
adding $1$ and $2$ we get
$2x = 16m/s$
$x = 8m/s$
and,
putting this in $1$
$y = 2m/s$
The speed of the bodies is $8 \mathrm{~m} / \mathrm{s}^2$ and $2 \mathrm{~m} / \mathrm{s}^2$.

Speed of body is magnitude of velocity. So if velocity is only changing direction then speed is constant.
But if speed is changing then velocity have to be varying.
When a body starts moving the incident when $t=0$ then it have zero velocity but some acceleration is there.
Object moving with constant velocity has zero acceleration but non zero speed.

The average speed of motion is $2.4\ m/s.$
Average Speed $=\frac{ \text{(Total distance covered throughout the journey)}}{\text{(Total time taken for the journey)}}$
Average Speed $=\frac{(10\text{m}+20\text{m}+30\text{m})}{(5\text{s}+10\text{s}+10\text{s})}$
Average Speed $=\frac{60\text{m}}{25\text{s}}=2.4\text{m/s}$

Let Partical start from A. then after covering $\frac{3}{4}$ of circle it would be at B.
So, Distence covred.
$1\text{ circle}=2\pi\text{r}\text{ units}.$
$\frac{3}{4}\text{ circle}=\frac{3}{4}\times2\pi\text{r}=\frac{3\pi\text{r}}{2}\text{units}$
Displacement:
Here, displacement will be line joining A to B.
So, by pythagoras theorem.
$\text{AB}=\sqrt{\text{r}^2+\text{r}^2}$
$=\text{r}\sqrt{2}$
So, Ratio of distance and displace will be
$\frac{3\pi\text{r}}{2}\times\frac{1}{\text{r}\sqrt{2}}=3\times\frac{22}{7}\times\frac{1}{2}\times\frac{1}{\sqrt{2}}=\frac{33}{7\sqrt{2}}$
So, ratio will be $3\pi:2\sqrt{2}$
or, simplified $33:7\sqrt{3}$

The ball initially will have the same velocity as that of the balloon.
Using the equations of motion, taking velocity to be positive upwards and negative downwards.
$v = u + at$
$v = 2 −g(1)$
$v = 7.8m/ s$

By the first eqn of motion
$v = u + at$ Put the given values
$v = 10 + 2 \times 4$
$u = 18\ m/ sec$