Explanation:

Explanation:

Choose the positive direction of \(x\) -axis to be from south to north. Then Velocity of train \(v_{T}=+10 m s ^{-1}\)
Velocity of parrot \(v_{P}=-5 m s ^{-1}\) Relative velocity of parrot with respect to train
\(=v_{P}-v_{T}=\left(-5 ms ^{-1}\right)-\left(+10 ms ^{-1}\right)=-15 m s ^{-1}\)
i.e. parrot appears to move with a speed of \(15 m s ^{-1}\) from north to south
\(\therefore\) Time taken by parrot to cross the train
\(=\frac{150 m }{15 ms ^{-1}}=10 s\)

Explanation:

Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)

Explanation:

\(\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}\)
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)

Explanation:

Average velocity \(=\frac{\text { Displacement }}{\text { Ttme tnterval }}\)
\(\bar{v}=\frac{+360 \mathrm{m}}{18 \mathrm{s}}=+20 \mathrm{m} \mathrm{s}^{-1}\)
\(\text {Average speed }=\frac{\text { Path length }}{\text { Tune truterval }}\)
\(=\frac{360 \mathrm{m}}{18 \mathrm{s}}=20 \mathrm{m} \mathrm{s}^{-1}\)

Explanation:

Explanation:

Choose the positive direction of \(x\) -axis to be from south to north. Then Velocity of train \(v_{T}=+10 m s ^{-1}\)
Velocity of parrot \(v_{P}=-5 m s ^{-1}\) Relative velocity of parrot with respect to train
\(=v_{P}-v_{T}=\left(-5 ms ^{-1}\right)-\left(+10 ms ^{-1}\right)=-15 m s ^{-1}\)
i.e. parrot appears to move with a speed of \(15 m s ^{-1}\) from north to south
\(\therefore\) Time taken by parrot to cross the train
\(=\frac{150 m }{15 ms ^{-1}}=10 s\)

Explanation:

Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)

Explanation:

\(\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}\)
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)

Explanation:

Average velocity \(=\frac{\text { Displacement }}{\text { Ttme tnterval }}\)
\(\bar{v}=\frac{+360 \mathrm{m}}{18 \mathrm{s}}=+20 \mathrm{m} \mathrm{s}^{-1}\)
\(\text {Average speed }=\frac{\text { Path length }}{\text { Tune truterval }}\)
\(=\frac{360 \mathrm{m}}{18 \mathrm{s}}=20 \mathrm{m} \mathrm{s}^{-1}\)

Explanation:

Explanation:

Choose the positive direction of \(x\) -axis to be from south to north. Then Velocity of train \(v_{T}=+10 m s ^{-1}\)
Velocity of parrot \(v_{P}=-5 m s ^{-1}\) Relative velocity of parrot with respect to train
\(=v_{P}-v_{T}=\left(-5 ms ^{-1}\right)-\left(+10 ms ^{-1}\right)=-15 m s ^{-1}\)
i.e. parrot appears to move with a speed of \(15 m s ^{-1}\) from north to south
\(\therefore\) Time taken by parrot to cross the train
\(=\frac{150 m }{15 ms ^{-1}}=10 s\)

Explanation:

Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)

Explanation:

\(\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}\)
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)

Explanation:

Average velocity \(=\frac{\text { Displacement }}{\text { Ttme tnterval }}\)
\(\bar{v}=\frac{+360 \mathrm{m}}{18 \mathrm{s}}=+20 \mathrm{m} \mathrm{s}^{-1}\)
\(\text {Average speed }=\frac{\text { Path length }}{\text { Tune truterval }}\)
\(=\frac{360 \mathrm{m}}{18 \mathrm{s}}=20 \mathrm{m} \mathrm{s}^{-1}\)

Explanation:

Explanation:

Choose the positive direction of \(x\) -axis to be from south to north. Then Velocity of train \(v_{T}=+10 m s ^{-1}\)
Velocity of parrot \(v_{P}=-5 m s ^{-1}\) Relative velocity of parrot with respect to train
\(=v_{P}-v_{T}=\left(-5 ms ^{-1}\right)-\left(+10 ms ^{-1}\right)=-15 m s ^{-1}\)
i.e. parrot appears to move with a speed of \(15 m s ^{-1}\) from north to south
\(\therefore\) Time taken by parrot to cross the train
\(=\frac{150 m }{15 ms ^{-1}}=10 s\)

Explanation:

Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)

Explanation:

\(\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}\)
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)

Explanation:

Average velocity \(=\frac{\text { Displacement }}{\text { Ttme tnterval }}\)
\(\bar{v}=\frac{+360 \mathrm{m}}{18 \mathrm{s}}=+20 \mathrm{m} \mathrm{s}^{-1}\)
\(\text {Average speed }=\frac{\text { Path length }}{\text { Tune truterval }}\)
\(=\frac{360 \mathrm{m}}{18 \mathrm{s}}=20 \mathrm{m} \mathrm{s}^{-1}\)

Explanation:

Explanation:

Choose the positive direction of \(x\) -axis to be from south to north. Then Velocity of train \(v_{T}=+10 m s ^{-1}\)
Velocity of parrot \(v_{P}=-5 m s ^{-1}\) Relative velocity of parrot with respect to train
\(=v_{P}-v_{T}=\left(-5 ms ^{-1}\right)-\left(+10 ms ^{-1}\right)=-15 m s ^{-1}\)
i.e. parrot appears to move with a speed of \(15 m s ^{-1}\) from north to south
\(\therefore\) Time taken by parrot to cross the train
\(=\frac{150 m }{15 ms ^{-1}}=10 s\)

Explanation:

Time to travel \(81 \mathrm{m}\) is t sec.
Time to travel \(100 \mathrm{m}\) is \(\mathrm{t}+\frac{1}{2} \mathrm{sec}\)
\(81=\frac{1}{2} \times \mathrm{a} \times \mathrm{t}^{2} \quad \Rightarrow \mathrm{t}=9 \sqrt{\frac{2}{\mathrm{a}}}\)
\(100=\frac{1}{2} \times \mathrm{a} \times\left(\mathrm{t}+\frac{1}{2}\right)^{2} \Rightarrow \mathrm{t}+\frac{1}{2}=10 \sqrt{\frac{2}{\mathrm{a}}}\)
\(9 \sqrt{\frac{2}{a}}+\frac{1}{2}=10 \sqrt{\frac{2}{a}}\)
\(\frac{1}{2}=\sqrt{\frac{2}{a}}\)
\(a=8 \mathrm{m} / \mathrm{s}^{2}\)

Explanation:

\(\mathrm{x}=\sqrt{\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}}\)
Differentiating w.r.t. time
\(\frac{d x}{d t}=v=\frac{1}{2 \sqrt{a t^{2}+2 b t+c}} \times(2 a t+2 b)\)
\(\Rightarrow \mathrm{v}=\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\)
\(\Rightarrow \mathrm{vx}=\mathrm{at}+\mathrm{b}\)
Differentiating w.r.t. \(\mathrm{x}\)
\(\Rightarrow \frac{d v}{d x} \times x+v=a \times \frac{d t}{d x}\)
Multiply both side by v
\(\Rightarrow\left(v \frac{d v}{d x}\right) x+v^{2}=a\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\mathrm{v}^{2}\) [Here \(a'\) is acceleration]
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\mathrm{a}-\left(\frac{\mathrm{at}+\mathrm{b}}{\mathrm{x}}\right)^{2}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ax}^{2}-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{a}\left(\mathrm{at}^{2}+2 \mathrm{bt}+\mathrm{c}\right)-(\mathrm{at}+\mathrm{b})^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime} \mathrm{x}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{2}}\)
\(\Rightarrow \mathrm{a}^{\prime}=\frac{\mathrm{ac}-\mathrm{b}^{2}}{\mathrm{x}^{3}}\)
\(\therefore a^{\prime} \propto \frac{1}{x^{3}} \quad \therefore n=3\)

Explanation:

Average velocity \(=\frac{\text { Displacement }}{\text { Ttme tnterval }}\)
\(\bar{v}=\frac{+360 \mathrm{m}}{18 \mathrm{s}}=+20 \mathrm{m} \mathrm{s}^{-1}\)
\(\text {Average speed }=\frac{\text { Path length }}{\text { Tune truterval }}\)
\(=\frac{360 \mathrm{m}}{18 \mathrm{s}}=20 \mathrm{m} \mathrm{s}^{-1}\)