Given:
Time = 8 seconds
Formula: (i)
Distance travelled = Area under speed - Time graph
= Area of triangle
\(=\frac{1}{2}\times20\times8\)
= 80m

(ii) \(Acceleration=\frac{\triangle v}{\triangle t}=\frac{20}{8}=\frac{5}{2}\)
=2.5 m/s²

Given:
Time = 2 seconds
Maximum speed = 18 km/hr
Here initial velocity u = 0,
\(v=\left( { v }_{ max } \right) =18\ km/h=18\times \frac { 5 }{ 18 } =5\ m/s\)
t₁=0, t₂ =2 seconds
\(a_{av}=\frac{v-u}{t_2-t_1}=\frac{\triangle v}{\triangle t}\)
so, \(a_{av}=\frac{5.0}{2}=2.5m/s^2\)

Given:
Speed of a particle v = 10t²m/s
Formula:
\(s=\int { v } dt\)
\(=\int _{ 2 }^{ 5 }{ 10{ t }^{ 2 }dt } =10\int _{ 2 }^{ 5 }{ { t }^{ 2 }dt } \)
\(=\frac { 10 }{ 3 } \left( { t }^{ 3 } \right) _{ 2 }^{ 5 }=\frac { 10 }{ 3 } \left( { 5 }^{ 3 }-{ 2 }^{ 3 } \right) =390m\)

Given:
t=3s
Distance s = Area of \(\triangle OAB\)

\(=\frac{1}{2}\times OA\times BA\)
\(=\frac{1}{2}\times 3\times 6=9m\)
If the speed varies with the time: Then \(v=\frac{ds}{dt}\Rightarrow ds=v\ dt\)
\(\Rightarrow \int { ds } =\int { v }\ dt\)
\(or\ s=\int { v }\ dt\)

Given: Total distance = s
Velocity of the girl for the first half distance = v₁
Velocity of the girl for the second half distance = v₂
Formula:
\(Time\ taken=\frac{Distance}{Velocity}\)
Time taken for the first half distance \(=\frac{\frac{s}{2}}{v_{1}}=\frac{s}{2v_1}\)
Time taken for the second half distance \(=\frac{\frac{s}{2}}{v_{2}}=\frac{s}{2v_2}\)
Total time taken \(=\frac{s}{2v_1}+\frac{s}{2v_2}\)
Average speed,\(v_{av}=\frac{s}{\frac{s}{2v_1}+\frac{s}{2v_2}}=\frac{2v_1v_2}{v_1+v_2}\)