QBManager

Motion in One Dimensions

141651 A train of $150 m$ length is going towards north direction at a speed of $10 {ms}^{- 1}$ . A parrot flies at speed at $5 {ms}^{- 1}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to:

1

12 s

2

8 s

3

15 s

4

10 s

5

5 s

Explanation:

D Given that,
Length of the train $= 150 m$
Speed of the train $(v_{t}) = 10 m / s$ (North direction)
Speed of parrot $(v_{P}) = 5 m / s$ (South direction)
We know that,
$Time = \frac{Distance}{Relative Velocity}$
$= \frac{Distance}{v_{t} - v_{p}}$
$= \frac{150}{10 - (- 5)} = \frac{150}{15} = 10 s$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141652 A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km h^{- 1}$ . If the muzzle speed of the bullet is $150 m s^{- 1}$ , with what speed does the bullet hit the thief's car?

1

90 m s^{- 1}

2

105 m s^{- 1}

3

110 m s^{- 1}

4

120 m s^{- 1}

Explanation:

B Given that,
Speed of the police van $(v_{police}) = 30 km / h = 30 \times \frac{5}{18}$
$= 8.33 m / s$
The muzzle speed of the bullet $(v_{bullet}) = 150 m / s$
Speed of the thief's car $(v_{thief}) = 192 km / h$
$= 192 \times \frac{5}{18} = 53.33 m / s$
The bullet is fired from a moving van, therefore resultant speed can be obtained as:
$= 150 + 8.33$
$= 158.33 m / s$
Both the vehicle are moving in the same direction. So, velocity of the bullet hits the thief's car is,
$v_{relative} = v_{bullet} - v_{thief}$
$= 158.33 - 53.33$
$= 105 m / s$

COMEDK 2017

Motion in One Dimensions

141653 A man of height $h$ walks in a straight path towards a lamp post of height $H$ with uniform velocity $u$ . Then the velocity of the edge of the shadow on the ground will be:

1

\frac{Hu}{(H - h)}

2

\frac{Hu}{(H + h)}

3

\frac{(H - h)}{Hu}

4

\frac{(H + h)}{Hu}

5

(\frac{H + h}{H - h}) u

Explanation:

A Let, same time the distance covered by man and edge of shadow be respectively $x$ and $y$ .
original image
Where, $CD =$ height of man $= h$
$AB =$ height of lamp-post $= H$
$y =$ length of shadow of the man
According to given figure,
$Δ ODC \approx Δ OAB$ ,
$\frac{CD}{BA} = \frac{OC}{OB}$
$\frac{h}{H} = \frac{y}{x + y} \Rightarrow \frac{H}{h} = \frac{x + y}{y}$
$\frac{H}{h} = \frac{x}{y} + \frac{y}{y} \Rightarrow \frac{H}{h} = \frac{x}{y} + 1$
$\frac{x}{y} = \frac{H}{h} - 1$
$\frac{H - h}{h} = \frac{x}{y}$
$∵$ Distance $=$ Speed $\times$ time
$x = u t$
and $(x + y) = vt$
$u t + y = v t$
$y = (v - u) t$
From equation (i), (ii) and (iii), we get-
$\frac{H - h}{h} = \frac{u t}{(v - u) t}$
$Hv - Hu - hv + hu = hu$
$v (H - h) = Hu$
$v = \frac{Hu}{(H - h)}$

UPSEE 2012

Motion in One Dimensions

141654 Two trains $A$ and $B$ of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 {kmh}^{- 1}$ in the same direction, with $A$ head of speed $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1 m / s^{2}$ . If after 50s, the guard of $B$ just brushes past the driver of $A$ , what was the original distance between them?

1

100 m

2

1150 m

3

1300 m

4

1250 m

Explanation:

D Given that,
For train A, $u_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
For train $B, u_{B} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
$a_{B} = 1 m / s^{2}, t = 50 s$
Distance covered by train B,
$s_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2}$
$s_{B} = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^{2}$
$s_{B} = 1000 + 1250$
$= 2250 m$
Distance covered by $train A, s_{A} = u_{A} \times t$
$= 20 \times 50$
$= 1000 m$
Hence, original distance between train A and B
$= s_{B} - s_{A} = 2250 - 1000 = 1250 m$

BCECE-2015

Motion in One Dimensions

141656 The angle made by $\vec{r} = 3 \vec{i} + 3 \vec{j}$ with the $x$ axis is

1

30^{\circ}

2

60^{\circ}

3

180^{\circ}

4

90^{\circ}

5

45^{\circ}

Explanation:

E If $\vec{r} = x \hat{i} + y \hat{j}$ then the angle made by position vector with the $x$ -axis is,
$\tan θ_{x} = \frac{y}{x}$
Here, $x = 3, y = 3$
$\tan θ_{x} = \frac{3}{3} = 1$
$\tan θ_{x} = \tan 45^{\circ}$
$θ_{x} = 45^{\circ}$

Kerala CEE 04.07.2022

Motion in One Dimensions

141651 A train of $150 m$ length is going towards north direction at a speed of $10 {ms}^{- 1}$ . A parrot flies at speed at $5 {ms}^{- 1}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to:

1

12 s

2

8 s

3

15 s

4

10 s

5

5 s

Explanation:

D Given that,
Length of the train $= 150 m$
Speed of the train $(v_{t}) = 10 m / s$ (North direction)
Speed of parrot $(v_{P}) = 5 m / s$ (South direction)
We know that,
$Time = \frac{Distance}{Relative Velocity}$
$= \frac{Distance}{v_{t} - v_{p}}$
$= \frac{150}{10 - (- 5)} = \frac{150}{15} = 10 s$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141652 A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km h^{- 1}$ . If the muzzle speed of the bullet is $150 m s^{- 1}$ , with what speed does the bullet hit the thief's car?

1

90 m s^{- 1}

2

105 m s^{- 1}

3

110 m s^{- 1}

4

120 m s^{- 1}

Explanation:

B Given that,
Speed of the police van $(v_{police}) = 30 km / h = 30 \times \frac{5}{18}$
$= 8.33 m / s$
The muzzle speed of the bullet $(v_{bullet}) = 150 m / s$
Speed of the thief's car $(v_{thief}) = 192 km / h$
$= 192 \times \frac{5}{18} = 53.33 m / s$
The bullet is fired from a moving van, therefore resultant speed can be obtained as:
$= 150 + 8.33$
$= 158.33 m / s$
Both the vehicle are moving in the same direction. So, velocity of the bullet hits the thief's car is,
$v_{relative} = v_{bullet} - v_{thief}$
$= 158.33 - 53.33$
$= 105 m / s$

COMEDK 2017

Motion in One Dimensions

141653 A man of height $h$ walks in a straight path towards a lamp post of height $H$ with uniform velocity $u$ . Then the velocity of the edge of the shadow on the ground will be:

1

\frac{Hu}{(H - h)}

2

\frac{Hu}{(H + h)}

3

\frac{(H - h)}{Hu}

4

\frac{(H + h)}{Hu}

5

(\frac{H + h}{H - h}) u

Explanation:

A Let, same time the distance covered by man and edge of shadow be respectively $x$ and $y$ .
original image
Where, $CD =$ height of man $= h$
$AB =$ height of lamp-post $= H$
$y =$ length of shadow of the man
According to given figure,
$Δ ODC \approx Δ OAB$ ,
$\frac{CD}{BA} = \frac{OC}{OB}$
$\frac{h}{H} = \frac{y}{x + y} \Rightarrow \frac{H}{h} = \frac{x + y}{y}$
$\frac{H}{h} = \frac{x}{y} + \frac{y}{y} \Rightarrow \frac{H}{h} = \frac{x}{y} + 1$
$\frac{x}{y} = \frac{H}{h} - 1$
$\frac{H - h}{h} = \frac{x}{y}$
$∵$ Distance $=$ Speed $\times$ time
$x = u t$
and $(x + y) = vt$
$u t + y = v t$
$y = (v - u) t$
From equation (i), (ii) and (iii), we get-
$\frac{H - h}{h} = \frac{u t}{(v - u) t}$
$Hv - Hu - hv + hu = hu$
$v (H - h) = Hu$
$v = \frac{Hu}{(H - h)}$

UPSEE 2012

Motion in One Dimensions

141654 Two trains $A$ and $B$ of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 {kmh}^{- 1}$ in the same direction, with $A$ head of speed $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1 m / s^{2}$ . If after 50s, the guard of $B$ just brushes past the driver of $A$ , what was the original distance between them?

1

100 m

2

1150 m

3

1300 m

4

1250 m

Explanation:

D Given that,
For train A, $u_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
For train $B, u_{B} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
$a_{B} = 1 m / s^{2}, t = 50 s$
Distance covered by train B,
$s_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2}$
$s_{B} = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^{2}$
$s_{B} = 1000 + 1250$
$= 2250 m$
Distance covered by $train A, s_{A} = u_{A} \times t$
$= 20 \times 50$
$= 1000 m$
Hence, original distance between train A and B
$= s_{B} - s_{A} = 2250 - 1000 = 1250 m$

BCECE-2015

Motion in One Dimensions

141656 The angle made by $\vec{r} = 3 \vec{i} + 3 \vec{j}$ with the $x$ axis is

1

30^{\circ}

2

60^{\circ}

3

180^{\circ}

4

90^{\circ}

5

45^{\circ}

Explanation:

E If $\vec{r} = x \hat{i} + y \hat{j}$ then the angle made by position vector with the $x$ -axis is,
$\tan θ_{x} = \frac{y}{x}$
Here, $x = 3, y = 3$
$\tan θ_{x} = \frac{3}{3} = 1$
$\tan θ_{x} = \tan 45^{\circ}$
$θ_{x} = 45^{\circ}$

Kerala CEE 04.07.2022

Motion in One Dimensions

141651 A train of $150 m$ length is going towards north direction at a speed of $10 {ms}^{- 1}$ . A parrot flies at speed at $5 {ms}^{- 1}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to:

1

12 s

2

8 s

3

15 s

4

10 s

5

5 s

Explanation:

D Given that,
Length of the train $= 150 m$
Speed of the train $(v_{t}) = 10 m / s$ (North direction)
Speed of parrot $(v_{P}) = 5 m / s$ (South direction)
We know that,
$Time = \frac{Distance}{Relative Velocity}$
$= \frac{Distance}{v_{t} - v_{p}}$
$= \frac{150}{10 - (- 5)} = \frac{150}{15} = 10 s$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141652 A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km h^{- 1}$ . If the muzzle speed of the bullet is $150 m s^{- 1}$ , with what speed does the bullet hit the thief's car?

1

90 m s^{- 1}

2

105 m s^{- 1}

3

110 m s^{- 1}

4

120 m s^{- 1}

Explanation:

B Given that,
Speed of the police van $(v_{police}) = 30 km / h = 30 \times \frac{5}{18}$
$= 8.33 m / s$
The muzzle speed of the bullet $(v_{bullet}) = 150 m / s$
Speed of the thief's car $(v_{thief}) = 192 km / h$
$= 192 \times \frac{5}{18} = 53.33 m / s$
The bullet is fired from a moving van, therefore resultant speed can be obtained as:
$= 150 + 8.33$
$= 158.33 m / s$
Both the vehicle are moving in the same direction. So, velocity of the bullet hits the thief's car is,
$v_{relative} = v_{bullet} - v_{thief}$
$= 158.33 - 53.33$
$= 105 m / s$

COMEDK 2017

Motion in One Dimensions

141653 A man of height $h$ walks in a straight path towards a lamp post of height $H$ with uniform velocity $u$ . Then the velocity of the edge of the shadow on the ground will be:

1

\frac{Hu}{(H - h)}

2

\frac{Hu}{(H + h)}

3

\frac{(H - h)}{Hu}

4

\frac{(H + h)}{Hu}

5

(\frac{H + h}{H - h}) u

Explanation:

A Let, same time the distance covered by man and edge of shadow be respectively $x$ and $y$ .
original image
Where, $CD =$ height of man $= h$
$AB =$ height of lamp-post $= H$
$y =$ length of shadow of the man
According to given figure,
$Δ ODC \approx Δ OAB$ ,
$\frac{CD}{BA} = \frac{OC}{OB}$
$\frac{h}{H} = \frac{y}{x + y} \Rightarrow \frac{H}{h} = \frac{x + y}{y}$
$\frac{H}{h} = \frac{x}{y} + \frac{y}{y} \Rightarrow \frac{H}{h} = \frac{x}{y} + 1$
$\frac{x}{y} = \frac{H}{h} - 1$
$\frac{H - h}{h} = \frac{x}{y}$
$∵$ Distance $=$ Speed $\times$ time
$x = u t$
and $(x + y) = vt$
$u t + y = v t$
$y = (v - u) t$
From equation (i), (ii) and (iii), we get-
$\frac{H - h}{h} = \frac{u t}{(v - u) t}$
$Hv - Hu - hv + hu = hu$
$v (H - h) = Hu$
$v = \frac{Hu}{(H - h)}$

UPSEE 2012

Motion in One Dimensions

141654 Two trains $A$ and $B$ of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 {kmh}^{- 1}$ in the same direction, with $A$ head of speed $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1 m / s^{2}$ . If after 50s, the guard of $B$ just brushes past the driver of $A$ , what was the original distance between them?

1

100 m

2

1150 m

3

1300 m

4

1250 m

Explanation:

D Given that,
For train A, $u_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
For train $B, u_{B} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
$a_{B} = 1 m / s^{2}, t = 50 s$
Distance covered by train B,
$s_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2}$
$s_{B} = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^{2}$
$s_{B} = 1000 + 1250$
$= 2250 m$
Distance covered by $train A, s_{A} = u_{A} \times t$
$= 20 \times 50$
$= 1000 m$
Hence, original distance between train A and B
$= s_{B} - s_{A} = 2250 - 1000 = 1250 m$

BCECE-2015

Motion in One Dimensions

141656 The angle made by $\vec{r} = 3 \vec{i} + 3 \vec{j}$ with the $x$ axis is

1

30^{\circ}

2

60^{\circ}

3

180^{\circ}

4

90^{\circ}

5

45^{\circ}

Explanation:

E If $\vec{r} = x \hat{i} + y \hat{j}$ then the angle made by position vector with the $x$ -axis is,
$\tan θ_{x} = \frac{y}{x}$
Here, $x = 3, y = 3$
$\tan θ_{x} = \frac{3}{3} = 1$
$\tan θ_{x} = \tan 45^{\circ}$
$θ_{x} = 45^{\circ}$

Kerala CEE 04.07.2022

Motion in One Dimensions

141651 A train of $150 m$ length is going towards north direction at a speed of $10 {ms}^{- 1}$ . A parrot flies at speed at $5 {ms}^{- 1}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to:

1

12 s

2

8 s

3

15 s

4

10 s

5

5 s

Explanation:

D Given that,
Length of the train $= 150 m$
Speed of the train $(v_{t}) = 10 m / s$ (North direction)
Speed of parrot $(v_{P}) = 5 m / s$ (South direction)
We know that,
$Time = \frac{Distance}{Relative Velocity}$
$= \frac{Distance}{v_{t} - v_{p}}$
$= \frac{150}{10 - (- 5)} = \frac{150}{15} = 10 s$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141652 A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km h^{- 1}$ . If the muzzle speed of the bullet is $150 m s^{- 1}$ , with what speed does the bullet hit the thief's car?

1

90 m s^{- 1}

2

105 m s^{- 1}

3

110 m s^{- 1}

4

120 m s^{- 1}

Explanation:

B Given that,
Speed of the police van $(v_{police}) = 30 km / h = 30 \times \frac{5}{18}$
$= 8.33 m / s$
The muzzle speed of the bullet $(v_{bullet}) = 150 m / s$
Speed of the thief's car $(v_{thief}) = 192 km / h$
$= 192 \times \frac{5}{18} = 53.33 m / s$
The bullet is fired from a moving van, therefore resultant speed can be obtained as:
$= 150 + 8.33$
$= 158.33 m / s$
Both the vehicle are moving in the same direction. So, velocity of the bullet hits the thief's car is,
$v_{relative} = v_{bullet} - v_{thief}$
$= 158.33 - 53.33$
$= 105 m / s$

COMEDK 2017

Motion in One Dimensions

141653 A man of height $h$ walks in a straight path towards a lamp post of height $H$ with uniform velocity $u$ . Then the velocity of the edge of the shadow on the ground will be:

1

\frac{Hu}{(H - h)}

2

\frac{Hu}{(H + h)}

3

\frac{(H - h)}{Hu}

4

\frac{(H + h)}{Hu}

5

(\frac{H + h}{H - h}) u

Explanation:

A Let, same time the distance covered by man and edge of shadow be respectively $x$ and $y$ .
original image
Where, $CD =$ height of man $= h$
$AB =$ height of lamp-post $= H$
$y =$ length of shadow of the man
According to given figure,
$Δ ODC \approx Δ OAB$ ,
$\frac{CD}{BA} = \frac{OC}{OB}$
$\frac{h}{H} = \frac{y}{x + y} \Rightarrow \frac{H}{h} = \frac{x + y}{y}$
$\frac{H}{h} = \frac{x}{y} + \frac{y}{y} \Rightarrow \frac{H}{h} = \frac{x}{y} + 1$
$\frac{x}{y} = \frac{H}{h} - 1$
$\frac{H - h}{h} = \frac{x}{y}$
$∵$ Distance $=$ Speed $\times$ time
$x = u t$
and $(x + y) = vt$
$u t + y = v t$
$y = (v - u) t$
From equation (i), (ii) and (iii), we get-
$\frac{H - h}{h} = \frac{u t}{(v - u) t}$
$Hv - Hu - hv + hu = hu$
$v (H - h) = Hu$
$v = \frac{Hu}{(H - h)}$

UPSEE 2012

Motion in One Dimensions

141654 Two trains $A$ and $B$ of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 {kmh}^{- 1}$ in the same direction, with $A$ head of speed $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1 m / s^{2}$ . If after 50s, the guard of $B$ just brushes past the driver of $A$ , what was the original distance between them?

1

100 m

2

1150 m

3

1300 m

4

1250 m

Explanation:

D Given that,
For train A, $u_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
For train $B, u_{B} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
$a_{B} = 1 m / s^{2}, t = 50 s$
Distance covered by train B,
$s_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2}$
$s_{B} = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^{2}$
$s_{B} = 1000 + 1250$
$= 2250 m$
Distance covered by $train A, s_{A} = u_{A} \times t$
$= 20 \times 50$
$= 1000 m$
Hence, original distance between train A and B
$= s_{B} - s_{A} = 2250 - 1000 = 1250 m$

BCECE-2015

Motion in One Dimensions

141656 The angle made by $\vec{r} = 3 \vec{i} + 3 \vec{j}$ with the $x$ axis is

1

30^{\circ}

2

60^{\circ}

3

180^{\circ}

4

90^{\circ}

5

45^{\circ}

Explanation:

E If $\vec{r} = x \hat{i} + y \hat{j}$ then the angle made by position vector with the $x$ -axis is,
$\tan θ_{x} = \frac{y}{x}$
Here, $x = 3, y = 3$
$\tan θ_{x} = \frac{3}{3} = 1$
$\tan θ_{x} = \tan 45^{\circ}$
$θ_{x} = 45^{\circ}$

Kerala CEE 04.07.2022

Motion in One Dimensions

141651 A train of $150 m$ length is going towards north direction at a speed of $10 {ms}^{- 1}$ . A parrot flies at speed at $5 {ms}^{- 1}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to:

1

12 s

2

8 s

3

15 s

4

10 s

5

5 s

Explanation:

D Given that,
Length of the train $= 150 m$
Speed of the train $(v_{t}) = 10 m / s$ (North direction)
Speed of parrot $(v_{P}) = 5 m / s$ (South direction)
We know that,
$Time = \frac{Distance}{Relative Velocity}$
$= \frac{Distance}{v_{t} - v_{p}}$
$= \frac{150}{10 - (- 5)} = \frac{150}{15} = 10 s$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141652 A police van moving on a highway with a speed of $30 km h^{- 1}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $192 km h^{- 1}$ . If the muzzle speed of the bullet is $150 m s^{- 1}$ , with what speed does the bullet hit the thief's car?

1

90 m s^{- 1}

2

105 m s^{- 1}

3

110 m s^{- 1}

4

120 m s^{- 1}

Explanation:

B Given that,
Speed of the police van $(v_{police}) = 30 km / h = 30 \times \frac{5}{18}$
$= 8.33 m / s$
The muzzle speed of the bullet $(v_{bullet}) = 150 m / s$
Speed of the thief's car $(v_{thief}) = 192 km / h$
$= 192 \times \frac{5}{18} = 53.33 m / s$
The bullet is fired from a moving van, therefore resultant speed can be obtained as:
$= 150 + 8.33$
$= 158.33 m / s$
Both the vehicle are moving in the same direction. So, velocity of the bullet hits the thief's car is,
$v_{relative} = v_{bullet} - v_{thief}$
$= 158.33 - 53.33$
$= 105 m / s$

COMEDK 2017

Motion in One Dimensions

141653 A man of height $h$ walks in a straight path towards a lamp post of height $H$ with uniform velocity $u$ . Then the velocity of the edge of the shadow on the ground will be:

1

\frac{Hu}{(H - h)}

2

\frac{Hu}{(H + h)}

3

\frac{(H - h)}{Hu}

4

\frac{(H + h)}{Hu}

5

(\frac{H + h}{H - h}) u

Explanation:

A Let, same time the distance covered by man and edge of shadow be respectively $x$ and $y$ .
original image
Where, $CD =$ height of man $= h$
$AB =$ height of lamp-post $= H$
$y =$ length of shadow of the man
According to given figure,
$Δ ODC \approx Δ OAB$ ,
$\frac{CD}{BA} = \frac{OC}{OB}$
$\frac{h}{H} = \frac{y}{x + y} \Rightarrow \frac{H}{h} = \frac{x + y}{y}$
$\frac{H}{h} = \frac{x}{y} + \frac{y}{y} \Rightarrow \frac{H}{h} = \frac{x}{y} + 1$
$\frac{x}{y} = \frac{H}{h} - 1$
$\frac{H - h}{h} = \frac{x}{y}$
$∵$ Distance $=$ Speed $\times$ time
$x = u t$
and $(x + y) = vt$
$u t + y = v t$
$y = (v - u) t$
From equation (i), (ii) and (iii), we get-
$\frac{H - h}{h} = \frac{u t}{(v - u) t}$
$Hv - Hu - hv + hu = hu$
$v (H - h) = Hu$
$v = \frac{Hu}{(H - h)}$

UPSEE 2012

Motion in One Dimensions

141654 Two trains $A$ and $B$ of length $400 m$ each are moving on two parallel tracks with a uniform speed of $72 {kmh}^{- 1}$ in the same direction, with $A$ head of speed $B$ . The driver of $B$ decides to overtake $A$ and accelerates by $1 m / s^{2}$ . If after 50s, the guard of $B$ just brushes past the driver of $A$ , what was the original distance between them?

1

100 m

2

1150 m

3

1300 m

4

1250 m

Explanation:

D Given that,
For train A, $u_{A} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
For train $B, u_{B} = 72 km / h = 72 \times \frac{5}{18} = 20 m / s$
$a_{B} = 1 m / s^{2}, t = 50 s$
Distance covered by train B,
$s_{B} = u_{B} t + \frac{1}{2} a_{B} t^{2}$
$s_{B} = 20 \times 50 + \frac{1}{2} \times 1 \times (50)^{2}$
$s_{B} = 1000 + 1250$
$= 2250 m$
Distance covered by $train A, s_{A} = u_{A} \times t$
$= 20 \times 50$
$= 1000 m$
Hence, original distance between train A and B
$= s_{B} - s_{A} = 2250 - 1000 = 1250 m$

BCECE-2015

Motion in One Dimensions

141656 The angle made by $\vec{r} = 3 \vec{i} + 3 \vec{j}$ with the $x$ axis is

1

30^{\circ}

2

60^{\circ}

3

180^{\circ}

4

90^{\circ}

5

45^{\circ}

Explanation:

E If $\vec{r} = x \hat{i} + y \hat{j}$ then the angle made by position vector with the $x$ -axis is,
$\tan θ_{x} = \frac{y}{x}$
Here, $x = 3, y = 3$
$\tan θ_{x} = \frac{3}{3} = 1$
$\tan θ_{x} = \tan 45^{\circ}$
$θ_{x} = 45^{\circ}$

Kerala CEE 04.07.2022