362089
A boat is sailing at a velocity \(\left( {3\hat i + 4\hat j} \right)\) with respect to ground and water in river is flowing with a velocity \(\left( { - 3\hat i - 4\hat j} \right).\) Velocity of the boat with respect to water is:
1 \(8\hat j\)
2 \(5\sqrt 2 \)
3 \(6\hat i + 8\hat j\)
4 \( - 6\hat i - 8\hat j\)
Explanation:
\({\overrightarrow v _{{\rm{BW}}}} = {\overrightarrow v _{BG}} - {\overrightarrow v _{RG}} = 6\hat i + 8\hat j\)
PHXI04:MOTION IN A PLANE
362090
A boat having a speed of 5 \(km\)/\(hr\) in still water, crosses a river of width 1 \(km\) long the shortest possible path in 15 minutes. The speed of the river in \(km\)/\(hr\).
362091
A man is crossing a river flowing with velocity of 5 \(m\)/\(s\). He reaches a point directly across at a distance of 60 \(m\) in 5 sec. His velocity in still water should be
1 \(12\,m/s\)
2 \(13\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Let \(\theta \) is the angle made by the velocity of man w.r.to \(y\) - axis. The time for crossing is \(t = \frac{d}{{\sqrt {{v^2} - {u^2}} }}\) \( \Rightarrow \,5 = \frac{{60}}{{\sqrt {{v^2} - {5^2}} }} \Rightarrow \,v = 13\,m/s\)
PHXI04:MOTION IN A PLANE
362092
A swimmer crosses a flowing stream of width \(\omega \) to and fro time \({t_1}\). The time taken to cover the same distance up and down the stream is \({t_2}\). If \({t_3}\) is the time the swimmer would take to swim a distance \(2\omega \) in still water, then
1 \(t_1^2 = {t_2}{t_3}\)
2 \(t_2^2 = {t_1}{t_3}\)
3 \(t_3^2 = {t_1}{t_2}\)
4 \({t_3} = {t_1} + {t_2}\)
Explanation:
Let \(v\) be the river velocity and \(u\) the velocity of swimmer in still water. Then \({t_1} = 2\left( {\frac{W}{{\sqrt {{u^2} - {v^2}} }}} \right)\) \({t_2} = \frac{W}{{u + v}} + \frac{W}{{u - v}} = \frac{{2uW}}{{{u^2} - {v^2}}}\) and \({t_3} = \frac{{2W}}{u}\) Now we can see that \(t_1^2 = {t_2}\,{t_3}\)
362089
A boat is sailing at a velocity \(\left( {3\hat i + 4\hat j} \right)\) with respect to ground and water in river is flowing with a velocity \(\left( { - 3\hat i - 4\hat j} \right).\) Velocity of the boat with respect to water is:
1 \(8\hat j\)
2 \(5\sqrt 2 \)
3 \(6\hat i + 8\hat j\)
4 \( - 6\hat i - 8\hat j\)
Explanation:
\({\overrightarrow v _{{\rm{BW}}}} = {\overrightarrow v _{BG}} - {\overrightarrow v _{RG}} = 6\hat i + 8\hat j\)
PHXI04:MOTION IN A PLANE
362090
A boat having a speed of 5 \(km\)/\(hr\) in still water, crosses a river of width 1 \(km\) long the shortest possible path in 15 minutes. The speed of the river in \(km\)/\(hr\).
362091
A man is crossing a river flowing with velocity of 5 \(m\)/\(s\). He reaches a point directly across at a distance of 60 \(m\) in 5 sec. His velocity in still water should be
1 \(12\,m/s\)
2 \(13\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Let \(\theta \) is the angle made by the velocity of man w.r.to \(y\) - axis. The time for crossing is \(t = \frac{d}{{\sqrt {{v^2} - {u^2}} }}\) \( \Rightarrow \,5 = \frac{{60}}{{\sqrt {{v^2} - {5^2}} }} \Rightarrow \,v = 13\,m/s\)
PHXI04:MOTION IN A PLANE
362092
A swimmer crosses a flowing stream of width \(\omega \) to and fro time \({t_1}\). The time taken to cover the same distance up and down the stream is \({t_2}\). If \({t_3}\) is the time the swimmer would take to swim a distance \(2\omega \) in still water, then
1 \(t_1^2 = {t_2}{t_3}\)
2 \(t_2^2 = {t_1}{t_3}\)
3 \(t_3^2 = {t_1}{t_2}\)
4 \({t_3} = {t_1} + {t_2}\)
Explanation:
Let \(v\) be the river velocity and \(u\) the velocity of swimmer in still water. Then \({t_1} = 2\left( {\frac{W}{{\sqrt {{u^2} - {v^2}} }}} \right)\) \({t_2} = \frac{W}{{u + v}} + \frac{W}{{u - v}} = \frac{{2uW}}{{{u^2} - {v^2}}}\) and \({t_3} = \frac{{2W}}{u}\) Now we can see that \(t_1^2 = {t_2}\,{t_3}\)
362089
A boat is sailing at a velocity \(\left( {3\hat i + 4\hat j} \right)\) with respect to ground and water in river is flowing with a velocity \(\left( { - 3\hat i - 4\hat j} \right).\) Velocity of the boat with respect to water is:
1 \(8\hat j\)
2 \(5\sqrt 2 \)
3 \(6\hat i + 8\hat j\)
4 \( - 6\hat i - 8\hat j\)
Explanation:
\({\overrightarrow v _{{\rm{BW}}}} = {\overrightarrow v _{BG}} - {\overrightarrow v _{RG}} = 6\hat i + 8\hat j\)
PHXI04:MOTION IN A PLANE
362090
A boat having a speed of 5 \(km\)/\(hr\) in still water, crosses a river of width 1 \(km\) long the shortest possible path in 15 minutes. The speed of the river in \(km\)/\(hr\).
362091
A man is crossing a river flowing with velocity of 5 \(m\)/\(s\). He reaches a point directly across at a distance of 60 \(m\) in 5 sec. His velocity in still water should be
1 \(12\,m/s\)
2 \(13\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Let \(\theta \) is the angle made by the velocity of man w.r.to \(y\) - axis. The time for crossing is \(t = \frac{d}{{\sqrt {{v^2} - {u^2}} }}\) \( \Rightarrow \,5 = \frac{{60}}{{\sqrt {{v^2} - {5^2}} }} \Rightarrow \,v = 13\,m/s\)
PHXI04:MOTION IN A PLANE
362092
A swimmer crosses a flowing stream of width \(\omega \) to and fro time \({t_1}\). The time taken to cover the same distance up and down the stream is \({t_2}\). If \({t_3}\) is the time the swimmer would take to swim a distance \(2\omega \) in still water, then
1 \(t_1^2 = {t_2}{t_3}\)
2 \(t_2^2 = {t_1}{t_3}\)
3 \(t_3^2 = {t_1}{t_2}\)
4 \({t_3} = {t_1} + {t_2}\)
Explanation:
Let \(v\) be the river velocity and \(u\) the velocity of swimmer in still water. Then \({t_1} = 2\left( {\frac{W}{{\sqrt {{u^2} - {v^2}} }}} \right)\) \({t_2} = \frac{W}{{u + v}} + \frac{W}{{u - v}} = \frac{{2uW}}{{{u^2} - {v^2}}}\) and \({t_3} = \frac{{2W}}{u}\) Now we can see that \(t_1^2 = {t_2}\,{t_3}\)
362089
A boat is sailing at a velocity \(\left( {3\hat i + 4\hat j} \right)\) with respect to ground and water in river is flowing with a velocity \(\left( { - 3\hat i - 4\hat j} \right).\) Velocity of the boat with respect to water is:
1 \(8\hat j\)
2 \(5\sqrt 2 \)
3 \(6\hat i + 8\hat j\)
4 \( - 6\hat i - 8\hat j\)
Explanation:
\({\overrightarrow v _{{\rm{BW}}}} = {\overrightarrow v _{BG}} - {\overrightarrow v _{RG}} = 6\hat i + 8\hat j\)
PHXI04:MOTION IN A PLANE
362090
A boat having a speed of 5 \(km\)/\(hr\) in still water, crosses a river of width 1 \(km\) long the shortest possible path in 15 minutes. The speed of the river in \(km\)/\(hr\).
362091
A man is crossing a river flowing with velocity of 5 \(m\)/\(s\). He reaches a point directly across at a distance of 60 \(m\) in 5 sec. His velocity in still water should be
1 \(12\,m/s\)
2 \(13\,m/s\)
3 \(5\,m/s\)
4 \(10\,m/s\)
Explanation:
Let \(\theta \) is the angle made by the velocity of man w.r.to \(y\) - axis. The time for crossing is \(t = \frac{d}{{\sqrt {{v^2} - {u^2}} }}\) \( \Rightarrow \,5 = \frac{{60}}{{\sqrt {{v^2} - {5^2}} }} \Rightarrow \,v = 13\,m/s\)
PHXI04:MOTION IN A PLANE
362092
A swimmer crosses a flowing stream of width \(\omega \) to and fro time \({t_1}\). The time taken to cover the same distance up and down the stream is \({t_2}\). If \({t_3}\) is the time the swimmer would take to swim a distance \(2\omega \) in still water, then
1 \(t_1^2 = {t_2}{t_3}\)
2 \(t_2^2 = {t_1}{t_3}\)
3 \(t_3^2 = {t_1}{t_2}\)
4 \({t_3} = {t_1} + {t_2}\)
Explanation:
Let \(v\) be the river velocity and \(u\) the velocity of swimmer in still water. Then \({t_1} = 2\left( {\frac{W}{{\sqrt {{u^2} - {v^2}} }}} \right)\) \({t_2} = \frac{W}{{u + v}} + \frac{W}{{u - v}} = \frac{{2uW}}{{{u^2} - {v^2}}}\) and \({t_3} = \frac{{2W}}{u}\) Now we can see that \(t_1^2 = {t_2}\,{t_3}\)
