$$v^{2}=u^{2}+2 as \Rightarrow s=\frac{v^{2}-u^{2}}{2 a}$$ $$v=u+at \Rightarrow t=\frac{v-u}{a}$$ $$\bar{v}=\frac{s}{t}=\frac{(v+u)(v-u)}{2 a} \times \frac{a}{v-u}=\frac{v+u}{2}$$

Let the car accelerate at rate $(\alpha)$ for time $(t_1)$ then maximum velocity attained, $$v=0+\alpha t_{1}=\alpha t_{1}$$ Now, the car decelerates at a rate $(\beta)$ for time $((t-t_1))$ and finally comes to rest. Then, $$0=v-\beta\left(t-t_{1}\right)$$ $$\begin{aligned}& \therefore 0=\alpha t_{1}-\beta t+\beta t_{1} \\& \therefore t_{1}=\frac{\beta}{\alpha+\beta} t \\& \therefore v=\frac{\alpha \beta}{\alpha+\beta} t\end{aligned}$$

$$T=\frac{u-(-u)}{a}=\frac{2 u}{a}$$ As $(a=1 ~m / s^{2})$, $$T=2 u$$

Initial velocity is $(-u)$ and acceleration is $(+1)$. Time to stop is $(t_{stop} = u)$. The displacement at $(t_{stop})$ is $(s_1 = -u^2 + \frac{1}{2}u^2 = -\frac{1}{2}u^2)$. The distance covered is $(\frac{1}{2}u^2)$. The time to return to origin is $(t_{return} = 2u)$. The distance covered from $(t_{stop})$ to $(t_{return})$ is also $(\frac{1}{2}u^2)$. The total distance traveled before reaching origin is $(\frac{1}{2}u^2 + \frac{1}{2}u^2 = u^2)$.

The distance travelled in first 10 second, $$S_{1}=\frac{1}{2} g(10)^{2}=50 g$$ The distance travelled in first 20 second, $$S=\frac{1}{2} g(20)^{2}=200 g$$ $$S_{2}=S-S_{1}=200 g-50 g=150 g=3 S_{1}$$

$$v^{2}=u^{2}-2 as=0$$ $$\Rightarrow s \propto v^{2}$$ $$\frac{s_{1}}{s_{2}}=\left(\frac{v_{1}}{v_{2}}\right)^{2}=(3)^{2}=9$$

Distance travelled in first 5s: $(s = \frac{1}{2} a t^2)$. So $(20 = \frac{1}{2} a (5)^2)$, which means $(a = \frac{40}{25} = 1.6 \text{ m/s}^2)$. Distance travelled in first 10s: $(S_{10} = \frac{1}{2} a (10)^2 = \frac{1}{2} (1.6) (100) = 80 \text{ m})$. Distance travelled in next 5s = $(S_{10} - S_5 = 80 - 20 = 60 \text{ m})$.

First case: $$(\frac{v}{3})^{2}=v^{2}-(2 a \times 6)$$ $$12 a=\frac{8 v^{2}}{9}$$ $$v^{2}=\frac{12 \times 9}{8} a=\frac{27}{2} a$$ Second case: $$\begin{aligned}& 0=\left(\frac{v}{3}\right)^{2}-2 ax \\& v^{2}=18 ax\end{aligned}$$ From (i) and (ii) $$\begin{aligned}& 18 ax=\frac{27}{2} a \\& x=\frac{3}{4}=0.75 ~cm\end{aligned}$$

$$v=u+at$$ $$0=-4+a \times(0.1)$$ $$a=40 ~m / s^{2}$$ $$s=\frac{v^{2}}{2 a}=\frac{16}{80}=0.2 ~m$$

$$s_{10}=\frac{1}{2}(2)(10)^{2}=100 ~m$$ $$s_{7}=\frac{1}{2}(2)(7)^{2}=49 ~m$$ Distance travelled in last 3 sec is $$s_{10}-s_{7}=100-49=51 ~m$$

$$s=ut+\frac{1}{2}at^{2}$$ $$=0+\left(\frac{1}{2} \times 20 \times(8)^{2}\right)$$ $$=640 ~cm$$

Let A and B will meet after time t sec. it means the distance travelled by both will be equal. $$s_{A}=ut=40 t$$ $$s_{B}=\frac{1}{2}at^{2}=\frac{1}{2} \times 4 \times t^{2}=2 t^{2}$$ As $(s_{A}=s_{B})$ $$40t = 2t^2$$ $$t=20 s$$

$$v=u+at$$ Taking initial velocity $(u=18 ~m / s)$ and final velocity $(v=-30 ~m / s)$ with time $(\Delta t=2.4 ~s)$. $$-30=18+a(2.4)$$ $$a=\frac{-30-18}{2.4}=-\frac{48}{2.4}=-20 ~m / s^{2}$$ Magnitude of acceleration is $(20 ~m / s^{2})$.

The distance travelled in first 10 second, $$S_{1}=\frac{1}{2} g(10)^{2}=50 g$$ The distance travelled in first 20 second, $$S=\frac{1}{2} g(20)^{2}=200 g$$ $$S_{2}=S-S_{1}=200 g-50 g=150 g=3 S_{1}$$

$$v^{2}=u^{2}-2 as=0 \Rightarrow s \propto v^{2}$$ $$\frac{s_{1}}{s_{2}}=\left(\frac{v_{1}}{v_{2}}\right)^{2}=(3)^{2}=9$$

Relative velocity of A with respect to B is $(v_{rel} = v_A - v_B = 30 - 20 = 10 \text{ m/s})$. Relative acceleration of A with respect to B is $(a_{rel} = a_A - a_B = -2 - 0 = -2 \text{ m/s}^2)$. The minimum distance required to avoid collision is $(d_{min} = \frac{v_{rel}^2}{2|a_{rel}|} = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25 \text{ m})$. For no collision, $(d > 25 \text{ m})$. Given correct answer is option 2 (d > 125m), which suggests a different setup or parameter not explicitly stated in the problem as presented in the OCR text, or is incorrect. Based on the provided numbers and standard kinematics, $(d > 25 \text{ m})$.

$$a=\frac{27.5-0}{10}=2.75 ~m / s^{2}$$ $$s=ut+\frac{1}{2}at^{2}=(27.5 \times 10)+\left[\frac{1}{2} \times 2.75 \times 100\right]=275+137.5=412.5 ~m$$

$$(50)^{2}=(10)^{2}+2 as$$ $$2as =2500-100=2400$$ Now, $$v^{2}=(50)^{2}+2(-a)(-s)=(50)^{2}+2 as=2500+2400=4900$$ $$v=70 ~m / s$$

$$s=\frac{1}{2}at^{2} \Rightarrow s \propto t^{2}$$ $$\therefore \frac{t_{1}}{t_{2}}=\sqrt{\frac{s_{1}}{s_{2}}}=\sqrt{\frac{s / 4}{s}}=\frac{1}{2}$$ $$t_{1}=\frac{1}{2} \times 4=2 ~s$$

The distance travelled in the $(n^{th})$ second, $(s_{n}=\frac{1}{2} g(2 n-1))$. Given $(g=10 ~m / s^{2})$, so $(\frac{1}{2}g=5)$. $$25=5(2 n-1)$$ $$2 n-1=5$$ $$n=3$$ The height from which it falls is $(s=\frac{1}{2}gt^{2})$. Since it falls for $(n=3)$ seconds, $$s=\frac{1}{2} \times 10 \times 3^2 = 5 \times 9=45 ~m$$

$$s_{n}=u+\frac{1}{2}a(2 n-1)$$ For 5th second: $$24=u+\frac{a}{2}(2\times5-1) \Rightarrow 24=u+\frac{9 a}{2} \ldots (i)$$ For 15th second: $$44=u+\frac{a}{2}(2\times15-1) \Rightarrow 44=u+\frac{29 a}{2} \ldots (ii)$$ Subtracting (i) from (ii) gives $$20 = \frac{29a}{2} - \frac{9a}{2} = \frac{20a}{2} = 10a$$ $$a=2 ~m / s^{2}$$

The distance covered in the $(n^{th})$ second, starting from rest, is $(s_{n}=\frac{a}{2}(2n-1))$. Distance in the 5th second: $(s_{5th}=\frac{a}{2}(2\times5-1) = \frac{9a}{2})$. Distance covered in 5 seconds, starting from rest: $(S_{5}=\frac{1}{2}at^{2}=\frac{1}{2}a(5)^{2}=\frac{25a}{2})$. The ratio is $(\frac{s_{5th}}{S_5} = \frac{9a/2}{25a/2} = \frac{9}{25})$.

Slope of v-t curve = acceleration. Since curves pass through origin, initial velocity $(u=0)$. Displacement $(s = ut + \frac{1}{2}at^2)$. Since $(u=0)$, $(s = \frac{1}{2}at^2)$. For a given time interval $(t)$, $(s \propto a)$. Therefore, the ratio of displacement of X to that of Y is equal to the ratio of their accelerations: $(\frac{s_X}{s_Y} = \frac{a_X}{a_Y} = \frac{3}{2})$.

$$nh=\frac{1}{2}gt^{2} \quad nb=ut$$ From these two equation we get, $$\begin{aligned}& nh=\frac{g}{2}\left(\frac{nb}{u}\right)^{2} \\& n=\frac{2 hu^{2}}{gb^{2}}\end{aligned}$$

$$s=ut+\frac{1}{2}at^{2}$$ $$s_{5}=0+\frac{1}{2}a(5)^{2}$$ $$x=\frac{25 a}{2}$$ $$s_{10}=0+\frac{1}{2}a(10)^{2}=\frac{100 a}{2}=4 x$$ $$\Delta s=s_{10}-s_{5}=4 x-x=3 x$$

$$s \propto u^{2}$$ $$\frac{s_{1}}{s_{2}}=\left(\frac{u_{1}}{u_{2}}\right)^{2}=\left(\frac{50}{100}\right)^{2}=\frac{1}{4}$$ $$s_{2}=4 s_{1}=4 \times 6=24 ~m$$

First case: $$(\frac{v}{3})^{2}=v^{2}-(2 a \times 6)$$ $$12 a=\frac{8 v^{2}}{9}$$ $$v^{2}=\frac{12 \times 9}{8} a=\frac{27}{2} a$$ Second case: $$\begin{aligned}& 0=\left(\frac{v}{3}\right)^{2}-2 ax \\& v^{2}=18 ax\end{aligned}$$ From (i) and (ii) $$\begin{aligned}& 18 ax=\frac{27}{2} a \\& x=\frac{3}{4}=0.75 ~cm\end{aligned}$$

$$\vec{u}=-10 \hat{k}$$ $$\vec{s}=\vec{u} t+\frac{1}{2}\vec{a}t^{2}=[(-10 \hat{k}) \times 2]+\frac{1}{2} \times(5 \hat{i}-3 \hat{j}+4 \hat{k}) \times(2)^{2}$$ $$=(-20 \hat{k})+(10 \hat{i}-6 \hat{j}+8 \hat{k})=10 \hat{i}-6 \hat{j}-12 \hat{k}$$

Distance travelled in successive equal time intervals for a body starting from rest and moving with constant acceleration are in the ratio 1:3:5: .... . Let S be the distance in the first 4 sec. Then the distance in the next 4 sec will be 3S.

$$s_{10}=\frac{1}{2} \times(2) \times(10)^{2}=100 ~m$$ $$s_{7}=\frac{1}{2} \times(2) \times(7)^{2}=49 ~m$$ Distance travelled in last $(3 sec)$ $(s_{10}-s_7)$ $(100-49=51 ~m)$.

For first ball, $$s=\frac{1}{2}gt^{2}$$ $$125=\frac{1}{2} \times 10 \times t^{2}$$ $$t=5 ~s$$ For second ball, $(t_{1}=3 ~s)$ $$s_{1}=ut_{1}+\frac{1}{2}gt_{1}^{2}$$ $$125=3 u+\frac{1}{2} \times 10 \times 9$$ $$u=\frac{80}{3}=26.7 ~m / s$$

$$s_{1}=ut+\frac{1}{2}at^{2}$$ $$s_{2}=vt+\frac{1}{2}at^{2}=(u+at) t+\frac{1}{2}at^{2}$$ Equation (ii) - (i), $$s_{2}-s_{1}=at^{2}$$ $$a=\frac{s_{2}-s_{1}}{t^{2}}=\frac{65-40}{(5)^{2}}=1 ~m / s^{2}$$ From equation (i), we get, $$s_{1}=ut+\frac{1}{2}at^{2}$$ $$40=5 u+\left[\frac{1}{2} \times 1 \times 25\right]$$ $$5 u=40-12.5=27.5$$ $$\therefore u=5.5 ~m / s$$

$$v=u+at$$ Taking initial velocity $(u=18 ~m / s)$ and final velocity $(v=-30 ~m / s)$ with time $(\Delta t=2.4 ~s)$. $$-30=18+a(2.4)$$ $$a=\frac{-30-18}{2.4}=-\frac{48}{2.4}=-20 ~m / s^{2}$$ Magnitude of acceleration is $(20 ~m / s^{2})$.

The distance covered in the $(n^{th})$ second is given by $(s_{n}=u+\frac{a}{2}(2n-1))$. Given $(u=7 ~m / s)$, $(a=4 ~m / s^{2})$, and $(n=5)$. $$s_{5th}=7+\frac{4}{2}((2\times5)-1)=7+2(9)=7+18=25 ~m$$

As distance travelled in successive seconds differs by 2 m each, therefore acceleration is constant. $$a=2 ~m / s^{2}$$ The distance in $(n^{th})$ second is $(s_{n}=u+\frac{a}{2}(2 n-1))$. For 8th second: $$s_{8}=u+\frac{2}{2} \times(2 \times 8-1)$$ $$20=u+1 \times(16-1)=u+15$$ $$u=5 ~m / s$$ Since $(u=5 ~m / s)$ (not zero), and $(a=2 ~m / s^{2})$ (not zero), the body starts with an initial velocity and moves with uniform acceleration.