Science M.C.Q. [1 M]

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 $= 0.5m/s^2$