143716
A particle moves in a straight line with a velocity \(v\) and moves under retardation equal to \(\mathrm{k}\) times the distance described. What is the distance covered before it comes to rest?
143717
Rain is falling vertically with a speed of \(12 \mathrm{~ms}^{-}\) 1. A woman rides a bicycles with a speed of 12 \(\mathrm{ms}^{-1}\) in east to west direction. What is the direction in which she should hold her umbrella ?
1 \(30^{\circ}\), towards east
2 \(45^{\circ}\), towards east
3 \(30^{\circ}\), towards west
4 \(45^{\circ}\), towards west [KarnatakaCET-2020]
Explanation:
D Given, Velocity of rain \(\mathrm{v}_{\mathrm{r}}=12 \mathrm{~ms}^{-1}\) Velocity of woman \(\mathrm{v}_{\mathrm{w}}=12 \mathrm{~ms}^{-1}\) \(\mathrm{V}_{\mathrm{rw}} =\sqrt{\mathrm{v}_{\mathrm{w}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}}\) \(=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\) \(\sin \theta=\frac{\mathrm{v}_{\mathrm{w}}}{\mathrm{v}_{\mathrm{rw}}}=\frac{12}{12 \sqrt{2}}\) \(\theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}\) west Hence, woman should hold her umbrella in \(45^{\circ}\) west direction.
Motion in Plane
143718
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
1 east-north direction
2 west-north direction
3 south-east direction
4 none of these
Explanation:
B Let, \(\overrightarrow{\mathrm{v}}_{\mathrm{c}}=\) Velocity of car \(\overrightarrow{\mathrm{v}}_{\mathrm{t}}=\) Velocity of Train Both speed is same \(\tan \theta =\frac{-\mathrm{v}}{\mathrm{v}} \Rightarrow \tan \theta=-1=\tan 135^{\circ}\) \(\theta =135^{\circ}\) Hence, velocity of car w.r. to train is towards west-north direction.
ВСЕСЕ-2018
Motion in Plane
143719
A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) East-West direction. If the rain falls vertically with a speed of \(10 \mathrm{~ms}^{-1}\), then the direction in which he must hold his umbrella is
1 \(30^{\circ}\) with the vertical towards East
2 \(60^{\circ}\) with the vertical towards West
3 \(30^{\circ}\) with the vertical towards West
4 \(60^{\circ}\) with the vertical towards East
5 \(0^{\circ}\) with the vertical
Explanation:
B Given that, A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) EastWest direction Speed of rain \(=10 \mathrm{~ms}^{-1}\) \(\tan \theta=\frac{\mathrm{v}_{\mathrm{m}}}{\mathrm{v}_{\mathrm{r}}}=\frac{17.32}{10}=1.732\) \(\tan \theta=1.732\) \(\tan \theta=\sqrt{3} \quad[\because \sqrt{3}=1.732]\) \(\tan \theta=\tan 60^{\circ}\) \(\theta=60^{\circ}\) with vertical towards west.
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Motion in Plane
143716
A particle moves in a straight line with a velocity \(v\) and moves under retardation equal to \(\mathrm{k}\) times the distance described. What is the distance covered before it comes to rest?
143717
Rain is falling vertically with a speed of \(12 \mathrm{~ms}^{-}\) 1. A woman rides a bicycles with a speed of 12 \(\mathrm{ms}^{-1}\) in east to west direction. What is the direction in which she should hold her umbrella ?
1 \(30^{\circ}\), towards east
2 \(45^{\circ}\), towards east
3 \(30^{\circ}\), towards west
4 \(45^{\circ}\), towards west [KarnatakaCET-2020]
Explanation:
D Given, Velocity of rain \(\mathrm{v}_{\mathrm{r}}=12 \mathrm{~ms}^{-1}\) Velocity of woman \(\mathrm{v}_{\mathrm{w}}=12 \mathrm{~ms}^{-1}\) \(\mathrm{V}_{\mathrm{rw}} =\sqrt{\mathrm{v}_{\mathrm{w}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}}\) \(=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\) \(\sin \theta=\frac{\mathrm{v}_{\mathrm{w}}}{\mathrm{v}_{\mathrm{rw}}}=\frac{12}{12 \sqrt{2}}\) \(\theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}\) west Hence, woman should hold her umbrella in \(45^{\circ}\) west direction.
Motion in Plane
143718
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
1 east-north direction
2 west-north direction
3 south-east direction
4 none of these
Explanation:
B Let, \(\overrightarrow{\mathrm{v}}_{\mathrm{c}}=\) Velocity of car \(\overrightarrow{\mathrm{v}}_{\mathrm{t}}=\) Velocity of Train Both speed is same \(\tan \theta =\frac{-\mathrm{v}}{\mathrm{v}} \Rightarrow \tan \theta=-1=\tan 135^{\circ}\) \(\theta =135^{\circ}\) Hence, velocity of car w.r. to train is towards west-north direction.
ВСЕСЕ-2018
Motion in Plane
143719
A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) East-West direction. If the rain falls vertically with a speed of \(10 \mathrm{~ms}^{-1}\), then the direction in which he must hold his umbrella is
1 \(30^{\circ}\) with the vertical towards East
2 \(60^{\circ}\) with the vertical towards West
3 \(30^{\circ}\) with the vertical towards West
4 \(60^{\circ}\) with the vertical towards East
5 \(0^{\circ}\) with the vertical
Explanation:
B Given that, A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) EastWest direction Speed of rain \(=10 \mathrm{~ms}^{-1}\) \(\tan \theta=\frac{\mathrm{v}_{\mathrm{m}}}{\mathrm{v}_{\mathrm{r}}}=\frac{17.32}{10}=1.732\) \(\tan \theta=1.732\) \(\tan \theta=\sqrt{3} \quad[\because \sqrt{3}=1.732]\) \(\tan \theta=\tan 60^{\circ}\) \(\theta=60^{\circ}\) with vertical towards west.
143716
A particle moves in a straight line with a velocity \(v\) and moves under retardation equal to \(\mathrm{k}\) times the distance described. What is the distance covered before it comes to rest?
143717
Rain is falling vertically with a speed of \(12 \mathrm{~ms}^{-}\) 1. A woman rides a bicycles with a speed of 12 \(\mathrm{ms}^{-1}\) in east to west direction. What is the direction in which she should hold her umbrella ?
1 \(30^{\circ}\), towards east
2 \(45^{\circ}\), towards east
3 \(30^{\circ}\), towards west
4 \(45^{\circ}\), towards west [KarnatakaCET-2020]
Explanation:
D Given, Velocity of rain \(\mathrm{v}_{\mathrm{r}}=12 \mathrm{~ms}^{-1}\) Velocity of woman \(\mathrm{v}_{\mathrm{w}}=12 \mathrm{~ms}^{-1}\) \(\mathrm{V}_{\mathrm{rw}} =\sqrt{\mathrm{v}_{\mathrm{w}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}}\) \(=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\) \(\sin \theta=\frac{\mathrm{v}_{\mathrm{w}}}{\mathrm{v}_{\mathrm{rw}}}=\frac{12}{12 \sqrt{2}}\) \(\theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}\) west Hence, woman should hold her umbrella in \(45^{\circ}\) west direction.
Motion in Plane
143718
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
1 east-north direction
2 west-north direction
3 south-east direction
4 none of these
Explanation:
B Let, \(\overrightarrow{\mathrm{v}}_{\mathrm{c}}=\) Velocity of car \(\overrightarrow{\mathrm{v}}_{\mathrm{t}}=\) Velocity of Train Both speed is same \(\tan \theta =\frac{-\mathrm{v}}{\mathrm{v}} \Rightarrow \tan \theta=-1=\tan 135^{\circ}\) \(\theta =135^{\circ}\) Hence, velocity of car w.r. to train is towards west-north direction.
ВСЕСЕ-2018
Motion in Plane
143719
A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) East-West direction. If the rain falls vertically with a speed of \(10 \mathrm{~ms}^{-1}\), then the direction in which he must hold his umbrella is
1 \(30^{\circ}\) with the vertical towards East
2 \(60^{\circ}\) with the vertical towards West
3 \(30^{\circ}\) with the vertical towards West
4 \(60^{\circ}\) with the vertical towards East
5 \(0^{\circ}\) with the vertical
Explanation:
B Given that, A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) EastWest direction Speed of rain \(=10 \mathrm{~ms}^{-1}\) \(\tan \theta=\frac{\mathrm{v}_{\mathrm{m}}}{\mathrm{v}_{\mathrm{r}}}=\frac{17.32}{10}=1.732\) \(\tan \theta=1.732\) \(\tan \theta=\sqrt{3} \quad[\because \sqrt{3}=1.732]\) \(\tan \theta=\tan 60^{\circ}\) \(\theta=60^{\circ}\) with vertical towards west.
143716
A particle moves in a straight line with a velocity \(v\) and moves under retardation equal to \(\mathrm{k}\) times the distance described. What is the distance covered before it comes to rest?
143717
Rain is falling vertically with a speed of \(12 \mathrm{~ms}^{-}\) 1. A woman rides a bicycles with a speed of 12 \(\mathrm{ms}^{-1}\) in east to west direction. What is the direction in which she should hold her umbrella ?
1 \(30^{\circ}\), towards east
2 \(45^{\circ}\), towards east
3 \(30^{\circ}\), towards west
4 \(45^{\circ}\), towards west [KarnatakaCET-2020]
Explanation:
D Given, Velocity of rain \(\mathrm{v}_{\mathrm{r}}=12 \mathrm{~ms}^{-1}\) Velocity of woman \(\mathrm{v}_{\mathrm{w}}=12 \mathrm{~ms}^{-1}\) \(\mathrm{V}_{\mathrm{rw}} =\sqrt{\mathrm{v}_{\mathrm{w}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}}\) \(=\sqrt{(12)^{2}+(12)^{2}}=12 \sqrt{2} \mathrm{~ms}^{-1}\) \(\sin \theta=\frac{\mathrm{v}_{\mathrm{w}}}{\mathrm{v}_{\mathrm{rw}}}=\frac{12}{12 \sqrt{2}}\) \(\theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}\) west Hence, woman should hold her umbrella in \(45^{\circ}\) west direction.
Motion in Plane
143718
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is
1 east-north direction
2 west-north direction
3 south-east direction
4 none of these
Explanation:
B Let, \(\overrightarrow{\mathrm{v}}_{\mathrm{c}}=\) Velocity of car \(\overrightarrow{\mathrm{v}}_{\mathrm{t}}=\) Velocity of Train Both speed is same \(\tan \theta =\frac{-\mathrm{v}}{\mathrm{v}} \Rightarrow \tan \theta=-1=\tan 135^{\circ}\) \(\theta =135^{\circ}\) Hence, velocity of car w.r. to train is towards west-north direction.
ВСЕСЕ-2018
Motion in Plane
143719
A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) East-West direction. If the rain falls vertically with a speed of \(10 \mathrm{~ms}^{-1}\), then the direction in which he must hold his umbrella is
1 \(30^{\circ}\) with the vertical towards East
2 \(60^{\circ}\) with the vertical towards West
3 \(30^{\circ}\) with the vertical towards West
4 \(60^{\circ}\) with the vertical towards East
5 \(0^{\circ}\) with the vertical
Explanation:
B Given that, A man rides a bicycle with a speed of \(17.32 \mathrm{~ms}^{-1}\) EastWest direction Speed of rain \(=10 \mathrm{~ms}^{-1}\) \(\tan \theta=\frac{\mathrm{v}_{\mathrm{m}}}{\mathrm{v}_{\mathrm{r}}}=\frac{17.32}{10}=1.732\) \(\tan \theta=1.732\) \(\tan \theta=\sqrt{3} \quad[\because \sqrt{3}=1.732]\) \(\tan \theta=\tan 60^{\circ}\) \(\theta=60^{\circ}\) with vertical towards west.
