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
$v = 10 + 2 \times 4$
$u = 18\ m/ sec$

Velocity-time graph of an object moving with uniform velocity.
The slope of a Velocity–time graph of an object moving in rectilinear motion with uniform velocity is straight line and parallel to $x-$axis when velocity is taken along $y-$axis and time is taken along $x-$axis.

Using first equation of motion
$v = u + at,$
we have
$0 = 20 + (−5)t$
$\Rightarrow 5t = 20$
$\Rightarrow t = 4s$

We know that Speed $(v)$
$=\frac{\text{Distance}}{\text{Time}}$
Here, Distance = Circumference of the circular path
$=2\pi\text{r}$
$=\frac{2\times22}{7\times21}=132$
and, Speed
$=2\text{sec}$
This gives, Time
$=\frac{132}{2}=66\text{m/sec}$
Total time $=66\text{Sec}$
then time taken by the body to complete half revolution is:
$=\frac{66}{2}=33\text{sec}$

Displacement is the minimum distance between the initial position and final position.
After moving half circle ($A$ to $B$) displacement will be $AB$
$AB$ is diameter i.e. $2r$

Train Length $= 150m$
Bridge length $= 400m$
Distance covered by train to cross bridge = Train Length + Bridge length
$= 150 + 400$
$= 550m$
$\text{Speed of train}=\frac{\text{Distance}}{\text{Time}}$
So,
$\text{Time}=\frac{\text{Distance}}{\text{Speed of train}}$
$=\frac{550}{50}$
$=11\text{sec}$
Time taken by the train to cross a bridge is $11sec.$

Acceleration of the moving object is $4 \mathrm{~m} / \mathrm{s}^2$
Initial velocity is $20\ m/s$.
Time taken is $2$ sec.
By applying $1^{\text {st }}$ equation of motion we get,$v = u + at$
$= 20 + 2(4)$
$= 28\ m/s$

Given, after half the circle, the particle will reach the diametrically opposite point i.e., from point A to point $6$. And we know displacement is shortest path between initial and final point.
Displacement after half circle $= AB = OA + OB$ [v Given, $OA $and $06 = r]$
Hence, the displacement after half circle is $2r.$

Velocity of man $= 10\ m/s$
$s = 48m$
Acceleration of man with respect to bus = Acceleration of man - Acceleration of bus
$= 0 - (1) = - 1m/s$
Applying second equation of motion,
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^{2}$
$48=10\text{t}-\frac{1}{2}\text{t}^{2}$
solving we get,
$t = 12sec$ or $t = 8sec$
therefore the minimum time is $8$ seconds.

Time required by $Q$ to move around $P$
$=$ Time taken by $Q$ for for one complete round - Time taken by $P$ for one complete round
$= 5min - 2min$
$= 3min.$

The slope of displacement - time graph gives us the velocity.
From the given graph, we see that the slope at $t=2$ is $0.$

Final velocity $= 72km/h = 20m/s$
Initial velocity is $36km/h = 10m/s$
Time taken is $10$ sec. So,
Acceleration $=\frac{20-10}{10}$
$=1\text{m/s}^2$

Displacement is the difference between the final and initial position of a body. It is a vector quantity and is independent of the path taken. So, for the movement of half of a circle, the displacement is $2r$, where r is the radius of the circular path.

For the first $40 m$
Using the third equation of motion: $\mathrm{v}^2-\mathrm{u}^2=2$ as
where: $\mathrm{s}=$
distance covered $=40 \mathrm{mv}=$
final velocity $=? \mathrm{~m} / \mathrm{su}=$
initial velocity $=0 \mathrm{~m} / \mathrm{sa}=$
acceleration $=1 \mathrm{~m} / \mathrm{s}^2$
$ v^2-u^2=2 a s$
$v^2-0=2 \times 1 \times 40$
$v^2=80$
$v=8.94 \mathrm{~m} / \mathrm{s}$
Using the first equation of motion $v=u+$ at
where:
$ \mathrm{v}=\text { final velocity }=8.94 \mathrm{~m} / \mathrm{s}$
$\mathrm{u}=\text { initial velocity }=0 \mathrm{~m} / \mathrm{sa}=\text { acceleration }=1 \mathrm{~m} / \mathrm{s}^2$
$\mathrm{t}_1=\text { time }=? \mathrm{~s}$
$\mathrm{v}=\mathrm{u}+\mathrm{at}_1$
$8.94=0+1 \times \mathrm{t}_1$
$\mathrm{t}_1=8.94 \mathrm{~s}$
Here its given the speed is uniform not velocity, therefore acceleration will be $1 \mathrm{~m} / \mathrm{s}^2$ and speed will be $8.94 \mathrm{~m} / \mathrm{s}$ everywhere.
Using the third equation of motion: $v^2-u^2=2$ as
where
$\mathrm{s}=\text { distance covered }$
$=60 \mathrm{mv}=\text { final velocity }=? ? \mathrm{~m} / \mathrm{su}$
$=\text { initial velocity }=8.94 \mathrm{~m} / \mathrm{sa}$
$=\text { acceleration }=1 \mathrm{~m} / \mathrm{s}^2$
$\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{as}$
$\mathrm{v}^2-(8.94)^2=2 \times 1 \times 60$
$\mathrm{v}^2-80=120$
$\mathrm{v}^2=120+80=200$
$v=14.14 \mathrm{~m} / \mathrm{s}$ Using the first equation of motion
$\mathrm{v}=\mathrm{u}+\mathrm{at}$
where:
$\mathrm{v}=\text { final velocity }=14.14 \mathrm{~m} / \mathrm{s}$
$\mathrm{u}=\text { initial velocity }=8.94 \mathrm{~m} / \mathrm{sa}=\text { acceleration }=1 \mathrm{~m} / \mathrm{s}^2$
$\mathrm{t}_2=\text { time }=? \mathrm{~s}$
$\mathrm{v}=\mathrm{u}+\mathrm{at}_2$
$14.14=8.94+1 \times \mathrm{t}_2$
$\mathrm{t}_2=14.14-8.94$
$\mathrm{t}_2=5.2 \text { secs. }$

We can convert $0.06\ m/s$ as:
$=\frac{0.06(3600)}{1000}\text{km/hr}$
$=0.216\text{km/hr}$
So, the answer is $0.216\ km/hr.$

A body is thrown vertically upward.
at maximum height velocity of body $= 0$
use kinematics formula ,
$v^2 = u^2 + 2as$
Here,
$v = 0$
$a = -g$
$s = H$
$0 = u^2 - 2gH$
$\text{H}=\frac{\text{u}^2}{2\text{g}}$
Hence, maximum height attained by body $\frac{\text{u}^2}{2\text{g}}$

If total distance is $S$ then
Average speed
$=\frac{\text{Total distance}}{\text{time}}$
$=\frac{\text{S}}{(\text{t}_{1}+\text{t}_{2})}$
$=\text{S}\ \frac{\text{S}}{(3\times10)}+\frac{\text{S}}{(4\times20)}+\frac{5\text{S}}{(12\times40)}$
$\therefore\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
$=\frac{1}{\big(30+1\frac{1}{80}+\frac{5}{480}\big)}=17.7\text{kmph}$

The term ''retardation'' means negative acceleration.Initial velocity $= 10m/s$
Final velocity $= 0m/s$
Time taken $= 20s$
Acceleration $=\frac{0-10}{20}$
$\Rightarrow $ Acceleration $= -0.5m/s^2$
$\Rightarrow $ Retardation $= 0.5m/s^2$

We have,
Initial velocity $= 20m/s$
Final velocity $= 0m/s$
Time taken $= 5 sec.$
So,
Deceleration = -Acceleration
$= \frac{-(0 - 20)}{5}$
$= 4\text{m/s}^2$

$\text{t}_\text{a}=\frac{\text{u}}{\text{g}}=10\text{s}$
same amount of time it will take to fall down

$\text{S}_{\text{B}}=70\text{t}-\frac{1}{2}(20)\text{t}^{2}$
$\text{S}_{\text{A}}=60\text{t}$
$\text{S}_{\text{B}}-\text{S}_{\text{A}}=2.5\text{km}$
$70\text{t}-10\text{t}^{2}-60\text{t}=2.5$
$\Rightarrow10\text{t}^{2}-10\text{t}+2.5=0$
$\Rightarrow 4\text{t}^{2}-4\text{t}+1=0$
$\Rightarrow(2\text{t}-1)^{2}=0$
$\Rightarrow\text{t}=\frac{1}{2}\text{hr}=30\text{min}.$

1. Drops a stone from the top of the building.
2. Simultaneously, starts a clock.
3. Stops the clock as the stone lands on the ground.
4. Follows the relation $\text{h}=\frac{1}{2}\text{gt}^2.$

3. Distance and time are directly proportional when the velocity is constant.

Explanation:

$\text{Speed}=\frac{\text{Time}}{\text{Distance}} $

For uniform velocity, Distance ∝ Time

1. Constant speed

Explanation:

When the distance an object travels is directly proportional to time, it is said to travel with Constant Speed. 

$\text{Speed}=\frac{\text{distance travelled}}{\text{time}}$

1. $\frac{\text{m}}{\text{s}}^{2}$

Explanation:

Si unit of retardation is same as that of acceleration as $\frac{\text{m}}{\text{s}}^{2}$