let the first particle takes time t to reach a position co-ordinate 15 m below the second particle. Obviously, the second body has fallen under gravity for a time (t -1) s only. Hence
15 = \({ y }_{ 1 }-{ y }_{ 2 }=\frac { 1 }{ 2 } gt^{ 2 }-\frac { 1 }{ 2 } g(t-1)^{ 2 }=gt-\frac { g }{ 2 } \)
or 15 = 10t - 5
\(\Rightarrow t=\frac { 15+5 }{ 10 } =2s.\) 

(i) Velocity attained by a particle after time t:
Let dt: be the change in velocity of the particle in time dt. Therefore, the acceleration of the particle is given by
\(a=\frac{dv}{dt} \ or \ dv=a \ dt\)
By integrating both sides, we get
\(\int{dv}= \int{a dt}\)
or \(\int{dv}= a \int{ dt}\)
or    v = at + k   --- (i)
where k is constant of integration.
when t = 0, v = u
Putting these values in equation (i), we get
k = u
Now putting the value of k in equation (i), we get
v = u + at
(ii) Displacement of the particle after time t:
Let dx be the displacement of the particle in time dt. Therefore, the velocity of the particle is given by
\(v=\frac{dx}{dt}\ or \ dx=vdt\)
Since v  =u + at
∴ dx=(u+at)dt
Integrating both sides, we get
\(\int{dx}=\int{(u+at)}dt\)
or  \(\int{dx}=\int{u}dt+\int{at \ dt}\) 
\(x= u\int{dt}+a \int{t\ dt}\)     [∵ u and a are constants]
or \(x=ut+a \frac{t^{2}}{2}+k\)
where k is constant of proportionality
where t = 0, x = x₀
∴ from equation (ii), we get
\(x=x_{0}+ut+\frac{1}{2}at^{2}\)
or \(x-x_{0}=ut+\frac{1}{2}at^{2}\)
since  x-x₀= S, displacement of the particle in the time interval t.
S = \(ut+\frac{1}{2}at^{2}\)
(iii) Velocity attained by a particle after travelling a distance S:
We know, \(v=\frac{dx}{dt}\)
Multiplying and dividing R.H.S. by dv, we get
\(v=\frac{dx}{dt}.\frac{dv}{dv}=\frac{dx}{dv}.\frac{dv}{dt}\)
As \(\frac{dv}{dt}=a (acceleration)\)
∴ v = a\(\frac { dx }{ dv } \) or v dv=a dx
Integrating both sides, we get \(\int { v\ dv=\int { a\ dx=a\int { dx } } } \)
or = \(\frac { v^{ 2 } }{ 2 } \)ax+k
when x = 0,v = u
Then,from eqn.(i),k = \(\frac { u^{ 2 } }{ 2 } \)
Putting the value of k in eqn. (i), we get
\(\frac { v^{ 2 } }{ 2 } \)-ax+\(\frac { u^{ 2 } }{ 2 } \)
or \(\frac { v^{ 2 } }{ 2 } -\frac { u^{ 2 } }{ 2 } =ax\)
or v²-u²= 2ax
x = s, then
v²-u² = 2 aS.

For uniformly accelerated motion, we can derive some simple equations that relate displacement(x), time taken(t), Initial velocity(u), final velocity(v) and acceleration(a).
(i) Velocity attained after time t: The velocity-time graph for positive constant acceleration of a particle is shown in the figure.

Let u be the initial velocity of the particle at t = 0 and v is the final velocity of the particle after time t. Consider two points A and B on the curve corresponding to t = 0 and t = t respectively.
Draw BD perpendicular to time axis. Also draw AC perpendicular to BD.
ஃ OA = CD = u;
BC = (v - u) and OD = t
Now slope of v-t graph = acceleration (a)
∴ a = slope of v - t graph = tan θ = \(\frac{BC}{AC}=\frac{BC}{OD} \ \ \ [∵ AC=OD]\)
∴ \(a=\frac{v-u}{t}\)
or v - u = at
v = u + at
(ii) Distance travelled in time t:
Let  x₀ position of the particle at t = 0 from the origin.
x = position of the particle at t = t from the origin.
∴ (x - x₀) = S = distance travelled by the particle in the time interval (t - 0) = t
We know, distance travelled by a particle in the given time interval =  area under velocity-time graph 
∴ (x - x₀) = Area OABD (see fig. above)
= Area of trapezium OABD
= \(\frac{1}{2}\) [Sum of parallel sides x perpendicular distance between parallel side]
=\(\frac{1}{2}(OA+BD)\times AC=\frac{1}{2}(u+v)t\)
Since v = u + at
∴  (x- x₀) = \(\frac{1}{2}(OA+BD)\times AC = \frac{1}{2} (u+v) \times t\)
Since x - x₀= S
∴   S = ut + \(\frac{1}{2}\)at²
(iii) Velocity attained after travelling a distance S:
We know, distance travelled by a particle in time t is equal to the area under velocitytime graph. Therefore, the distance (S) travelled by a particle during time interval tis given by
S = Area under v - t graph (see fig.) or
S = area of trapenium OABD
= \(\frac{1}{2}\)(sum of parallel sides) x perpendicular distance between these parallel
or S = \(\frac{1}{2}\)(OA + BD) x AC    --- (i)
Now, acceleration, a = slope of v - t graph
or a = \(\frac{BC}{AC}=\frac{BD-CD}{AC}=\frac{v-u}{AC}\) 
or \(AC=(\frac{v-u}{a})\)     --- (ii)
Also OA = u and BD = v  --- (iii)
Using equations (ii) and (iii) in equation (i), we get
\(S=\frac{1}{2}(v+u)\frac{(v-u)}{a}=\frac{v^{2}-u^{2}}{2a}\)
or, v²-u² = 2aS

(a) Distance travelled by the particle between t = 0 s to 10 s
= area of \(\Delta \)DAB = \(\frac { 1 }{ 2 } \) base x height
=\(\frac { 1 }{ 2 } \)x 10 x 12 = 60m
\(\therefore \) Average speed of particle v_av = \(\frac { 60m }{ 10s } \)= 6 ms^-1
(b) The distance traversed by the particle between
t = 2s to t = 6 s
= distance from 2 to 5 s (s₁) + distance in 6th second (s₂)
Now, u = 0, t = 5, v = 12 ms^{-1
\(\therefore \)} Acceleration for 0 - 5 s, a =\(\frac { v-u }{ t } =\frac { 12-0 }{ 5 } \) ms^-2 = 2.4 ms^-2
∴ Distance covered from 2 to 5 s = distance covered in 5 s - distance covered in 2 s
S₁ =  \(\frac { 1 }{ 2 } a(5)^{ 2 }-\frac { 1 }{ 2 } a(2)^{ 2 }=\frac { 1 }{ 2 } \times 2.4\times \left[ (5)^{ 2 }-(2)^{ 2 } \right] \)
= 25.2 m.
For motion from 5 to 10 s, u = +12 ms^-1 and a = -2.4 ms^-2 and interval t = 5 s to t = 6 s means n = 1 for this motion.
∴ Distance covered in 6th second 52 = u + \(\frac { 1 }{ 2 } \)a (2n -1)
= 12 - \(\frac { 2.4 }{ 2 } \) (2 x 1-1 ) = 10.8 m
∴ Total distance covered from t = 2 s to 6 s = S1 + S2
= 25.2 + 10.8 = 36 m and average speed =\(\frac { 36M }{ (6-2)S } \) 9 ms^-1.

Here, u = 100 ms^-1,g =- 10ms^-1
At highest point, v=0
As     v = u + gt
\(\Rightarrow \)   0 = 100 - 10\(\times \)t
Time taken to reach highest point
 \(t=\frac { 100 }{ 10 } =10s\)
The ball will return to the ground at t = 20 s. Velocities of the ball at different instants of time will be as follows.
At t =0,\(v=100-10\times 0=100m{ s }^{ -1 }\)
At t =5 ,\(v=100-10\times 5=50m{ s }^{ -1 }\)
At t=10,\(v=100-10\times 10=0\)
At t=15,\(v=100-10\times 15=-50m{ s }^{ -1 }\)
At t = 20, \(v=100-10\times 20=-100m{ s }^{ -1 }\) 
The velocity time-graph will be as shown in figure.
 
(i) Maximum height attained by ball
= Area of \(\Delta AOB\)
=\(\frac { 1 }{ 2 } \times 10s\times 100m{ s }^{ -1 }=500m\)
(ii) Height attained after 15 s
= Area of \(\Delta AOB\) + Area of \(\Delta BCD\)
= 500 + (15-10)x (- 50)
= 500-125 = 375 m

Area  AEFG = AE x AG \(\Rightarrow \) 7 = 40 x AG
 AG = \(\frac { 7 }{ 40 } h\)
Area FGB gives the distance covered under retardation, it is (8.5 -7) km = 1.5km
Area of \(\Delta FGB=\frac { 1 }{ 2 } GB\times FG\Rightarrow GB=\frac { 2\times 1.5 }{ 40 } h=\frac { 3 }{ 40 } h\) 
Total time = \(\left( \frac { 7 }{ 40 } +\frac { 3 }{ 40 } \right) h=\frac { 1 }{ 4 } h\) 
Area of 
\(\Delta ACB=\frac { 1 }{ 2 } \times AB\times CD\) 
\(8.5=\frac { 1 }{ 2 } \times \frac { 1 }{ 4 } \times v\)
\( v\ =8.5\times 8km{ h }^{ -1 }=68km{ h }^{ -1 }\)

Given: t₁=8 hr, t₂=12 hr
Let, s be the distance between that two spots. Also assume that the velocity of the motor boat in still water is v and the velocity of flow of water is u.
Then, for downward journey,
\(\frac{\mathrm{s}}{\mathrm{t}_1}=\mathrm{v}+\mathrm{u}\)
For upward journey,
\(\frac{\mathrm{s}}{\mathrm{t}_2}=\mathrm{v}-\mathrm{u}\)
Adding equation (i) to (ii),
\( \frac{\mathrm{s}}{\mathrm{t}_1}+\frac{\mathrm{s}}{\mathrm{t}_2}=2 \mathrm{v} \)
\( 2 \mathrm{v}=\mathrm{s}\left[\frac{1}{8}+\frac{1}{12}\right]=s\left[\frac{5}{24}\right] \)
\(\mathrm{v}=\mathrm{s}\left[\frac{5}{48}\right] \Rightarrow \frac{\mathrm{s}}{\mathrm{v}}=\frac{48}{5}\)
Time required for boat to cover distance s in still water
\(\mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}\)
\(\therefore \mathrm{t}=\frac{48}{5}=9.6 \mathrm{hr}\)
Hence that proved

Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²

(i) Draw position-time graph for
(a) Accelerated motion  
(b) Retarded motion



(ii) Juggler throws the ball into the air with ball having initial velocity = u
At the highest point of its path velocity will be zero as at maximum height velocity =0
So final velocity=0
According to question juggler throws a ball each second so for "n" balls time taken to reach highest position (t)= 1/n
v=u+gt
0=u-g/n
-u=-g/n
u=g/n
S= ut-gt²/2
S=u/n-g/2n²
But u=g/n
S= g/n²-g/2n²
S=g/2n²

Taking South to North direction as the positive direction 
i.e., x-axis, we have \({ v }_{ A }=+54km/h=\frac { 54\times 1000 }{ 3600 } { ms }^{ -1 }=15ms^{ -1 }\) 
\(\\ { v }_{ B }=-90km/h\ \)
\(=\frac { 90\times 1000 }{ 3600 } { ms }^{ -1 }=-25{ ms }^{ -1 }\)
 Relative velocity of monkey with respect to train B
= 0- v_B= 0 + 25 = 25ms^-1
So, to an observer in train B, the Earth appears to move with a speed of 25 m/s from South to North.