Science Que-Ans (Each of 5 Mark )

1. Initial velocity, $u = 0 [$since displacement and time is zero $]$

2. Velocity after $100s,$ $\text{v}=\frac{\text{Displacement}}{\text{Time}} [$Here, displacement $=$ zero and time $=$ $100s]$

1. $\text{v}=\text{u}+\text{at}$
2. $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
3. $\text{v}^2=\text{u}^2+2\text{as}$

1. $v = u$
2. $s = ut$
3. $v^2 = u^2$

1. The distance covered from $A$ to $B,$

2. The speed of the body as it moves from $B$ to $C$ is zero because the distance travelled is zero.
3. The distance covered from $C$ to $D,$

1. From the graph we can infer that in the interval $OA,$ it’s almost a straight line with a positive slope. Therefore velocity is positive that is with increase in time distance also increases and thus acceleration is also positive as well as uniform.
2. In the interval $AB,$ the line has a negative slope as with increase in time distance decreases. Here velocity is negative and it shows retardation.
3. $BC$ represents more downfalls in the slope of the line which means velocity will become more negative and hence it shows retardation with greater magnitude than $AB.$
4. $CD$ shows that with increase in time distance is increasing again with a positive slope. Therefore velocity is positive which implies acceleration is also positive.

1. Total distance covered in going to the bookshop and coming back to the classroom $= 20 + 20 = 40m$

2. Average velocity $=\frac{\text{Total displacement}}{\text{Total time}}=\frac{0}{50}=0\text{m/s}$

1. In the first part, train travels at a speed of $30\ km/h$ for distance of $15\ km.$

2. In the second part, train travels at a speed of $50\ km/h$ for a distance of $75\ km.$

3. In the third part, train travels at a speed of $20\ km/h$ for a distance of $10\ km.$

1. Fina; velocity = $v = 2u = 2 \times 5 \times 10^4ms^{-1} = 10 \times 10^4ms^{-1}$​​​​​​​

2. Using $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

1. $OA$ represents uniform acceleration.
2. $AB$ represents constant speed.
3. $BC$ represents uniform retardation.
4. Acceleration of car from $O$ to $A =$ slope of line $OA$

5. Acceleration of car from $A$ to $B$ is zero as it has uniform speed.

6. Retardation of car from $B$ to $C =$ slope of line $BC.$

1. Initial speed of the car is $10\ km/h$
2. Maximum speed attained by the car is $35\ km/h$
3. $BC$ represents zero acceleration.
4. $CD$ represents varying retardation.
5. Distance travelled is given by the area enclosed within the curve. So,

1. Graph $(a)$ represents uniformly accelerating motion as it has a constant slope.
2. Graph $(b)$ represents a motion of constant speed.
3. Graph $(c)$ represents uniformly retarding motion as it has a constant negative slope.
4. Graph $(d)$ represents non-uniformly retarding as it has a varying slope.

1. Area of rectangle $OADC = OA × OC$

2. Area of triangle ABD $=\Big(\frac{1}{2}\Big)\times\text{Area of rectangle AEBD}$

1. Initial speed of the car is 10 km/h
2. Maximum speed attained by the car is 35km/h
3. BC represents zero acceleration.
4. CD represents varying retardation.
5. Distance travelled is given by the area enclosed within the curve. So,

Distance travelled = Area or trapezium + Area of rectangle

So distance travelled,

$= \Big(\frac{1}{2}\Big)(8+5)(25) + (8)(10)$

$= 242.5\text{km}$

1. OA represents uniform acceleration.
2. AB represents constant speed.
3. BC represents uniform retardation.
4. Acceleration of car from O to A = slope of line OA

$\text{a}=\frac{(40-0)}{(10-0)}\text{m/s}^2$

$=4\text{m/s}^2$

5. Acceleration of car from A to B is zero as it has uniform speed.

6. Retardation of car from B to C = slope of line BC.

$\text{a}=\frac{(40-0)}{(50-30)}\text{m/s}^2$

$=2\text{m/s}^2$

1. Area of rectangle OADC = OA × OC

= u × t

= ut

2. Area of triangle ABD $=\Big(\frac{1}{2}\Big)\times\text{Area of rectangle AEBD}$

$=\Big(\frac{1}{2}\Big)\times\text{AD}\times\text{BD}$

$=\Big(\frac{1}{2}\Big)\text{at}^2$

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

1. The distance covered from A to B,

= 3 - 0

= 3cm

Time taken to cover the distance from A to B

= 5 - 2

= 3s

Hence speed,

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

$=\frac{3}{3}\text{cm/s}$

$=1\text{cm/s}$

2. The speed of the body as it moves from B to C is zero because the distance travelled is zero.
3. The distance covered from C to D,

= 7 - 3

= 4cm

Time taken to cover the distance from C to D,

= 9 - 7

= 2s

Hence speed,

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

$=\frac{4}{2}\text{cm/s}$

$=21\text{cm/s}$

1. $\text{v}=\text{u}+\text{at}$

2. $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

3. $\text{v}^2=\text{u}^2+2\text{as}$

v = Final velocuty

a = Acceleration

t = Time

s = Distance

For an object moving with uniform velocity (velocity which is not changing with time), then acceleration a = 0.

1. v = u
2. s = ut
3. v² = u²

$\text{s}=\frac{(\text{Sum of parallel sides})\times\text{Height}}{2}$

$\text{s}=\frac{(\text{OA+CB})\times\text{OC}}{2}$

$\text{s}=\Big(\frac{\text{u+v}}{2}\Big)\times\text{t}\ ...(1)$

1. In the first part, train travels at a speed of 30 km/h for distance of 15km.

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

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

$\text{t}_1=\frac{15}{30}=\frac{1}{2}\text{h}$

2. In the second part, train travels at a speed of 50km/h for a distance of 75km.

$\text{t}_2=\frac{75}{50}=\frac{3}{2}\text{h}$

3. In the third part, train travels at a speed of 20km/h for a distance of 10km.

$\text{t}_3=\frac{10}{20}=\frac{1}{2}\text{h}$

Total distance covered = 15 + 75 + 10 = 100km

Average speed $=\frac{\text{Total distance covered}}{\text{Total time taken}}$

$=\frac{100}{\frac{5}{2}}=40\text{km/h}$

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2$

$\text{h}=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^{2}]$

$\text{h}=0+\frac{1}{2}\times10\times4=20\text{m}$

$\text{h}=\text{ut}+\frac{1}{2}\text{gt}^2=0\times2+\frac{1}{2}\times10\times(2)^2\ [\because\text{g}=10\text{m/ s}^2]$

$=0+\frac{1}{2}\times1=\times4=20\text{m}$

1. Fina; velocity = v = 2u = 2 × 5 × 10⁴ms^-1 = 10 × 10⁴ms^-1
   To find t, use v = u + at
   or  $\text{t}=\frac{\text{v-u}}{\text{a}}$
   $\Big(\frac{10\times10^2-5\times10^4}{10^4}\Big)=\frac{5\times10^4}{10^4}=5\text{s}$

2. Using $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
   $=(5\times10^4)\times5+\frac{1}{2}(10^4)\times(5)^2=25\times10^4+\frac{25}{2}\times10^4=37.5\times10^4\text{m}$

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

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

$\text{t}_1=\frac{100}{60}\text{h}$

$\text{t}_2=\frac{100}{40}\text{h}$

$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}$

$\text{a}=\Big[\frac{0-400}{2(50)}\Big]\text{m/s}^2$

$=\Big(-\frac{400}{100}\Big)\text{m/s}^2$

$=-4\text{m/s}^2$

$\text{v}=\sqrt{\text{u}^2+2\text{as}}$

$\text{v}=\sqrt{64+36}\text{m/s}$

$=\sqrt{100}\text{m/s}$

$=10\text{m/s}$

1. Total distance covered in going to the bookshop and coming back to the classroom = 20 + 20 = 40m

Total time taken = 25 + 25 = 50 sec

Average speed $=\frac{\text{Total distance}}{\text{Total time}}=\frac{40}{50}=0.8\text{m/s}$

2. Average velocity $=\frac{\text{Total displacement}}{\text{Total time}}=\frac{0}{50}=0\text{m/s}$

$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$

$100=0\times5+\frac{1}{2}\times\text{a}\times5\times5$

$100=0+\frac{25\text{a}}{2}$

$\text{a}=\frac{200}{25}=8\text{m/s}^2$