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Motion in One Dimensions

141206 The instantaneous velocity of a particle moving in a straight line is given as $v = α t + β t^{2}$ , where $α$ and $β$ are constants. The distance travelled by the particle between $1 s$ and $2 s$ is

1

3 α + 7 β

2

\frac{3}{2} α + \frac{7}{3} β

3

\frac{α}{2} + \frac{β}{3}

4

\frac{3}{2} α + \frac{7}{2} β

Explanation:

B The instantaneous velocity of a particle is given as -
$v = α t + β t^{2}$
$∵ v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t + β t^{2}$
$d x = (α t + β t^{2}) d t$
Integrating the above equation from, $t = 1$ to 2
$\int_{1}^{2} dx = \int_{1}^{2} [α t] dt + \int_{1}^{2} β t^{2} dt$
$[x]_{1}^{2} = \frac{[α t]_{1}^{2}}{2} + \frac{{[β t^{3}]}_{1}^{2}}{3}$
$[x]_{1}^{2} = \frac{α}{2} [2^{2} - 1^{1}] + \frac{β}{3} [2^{3} - 1^{3}]$
$[x]_{1}^{2} = \frac{α}{2} [4 - 1] + \frac{β}{3} [8 - 1]$
$Δ x = (x_{2} - x_{1}) = \frac{3 α}{2} + \frac{7 β}{3}$

JEE Main-25.07.2021

Motion in One Dimensions

141207 A time dependent force is acting on a body of which the velocity of body can be expressed as $α t^{3 / 2}$ . The displacement of the body between time $t = 2$ and $5 \sec$ is about.

1

20 α

2

22 α

3

6 α

4 0

Explanation:

A Velocity of the body is given as -
$v = α t^{3 / 2}$
To find out the displacement between time $(t) = 2$ second to 5 second
$v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t^{3 / 2}$
$dx = α t^{3 / 2} dt$
Integrating the above equation from $t = 2$ to $5 s -$
$\int_{2}^{5} dx = \int_{2}^{5} α t^{3 / 2} dt$
$[x_{5} - x_{2}] = α {[\frac{t^{5 / 2}}{5 / 2}]}_{2}^{5}$
$Δ x = α \times \frac{2}{5} [5^{2.5} - 2^{2.5}]$
$Δ x = α \times \frac{2}{5} [55.90 - 5.65]$
$Δ x = \frac{2 α}{5} [50.24]$
$Δ x = 20.09 α \approx 20 α$
$Δ x = 20 α$

Assam CEE-2021

Motion in One Dimensions

141208 An object is moving along $x$ -axis. If its position $x (t)$ (in meter) at any time $t$ (in $s$ ) is given by the following graph, then the object is at rest about
original image

1

t = - 1 s

2

t = 0 s

3

t = 1 s

4

t = 2 s

Explanation:

C An object is moving along $x$ -axis. If its position $x (t)$ in meter at any time $+ ($ in sec) is given by the following graph. Then the object is at rest about $t = 1 \sec$
Because at $t = 1 \sec$ position of object is not changing.
original image

TS EAMCET 28.09.2020

Motion in One Dimensions

141209 A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of $72 km / hr$ . The angle of inclination of the bike with horizontal plant will be

1

30^{\circ}

2

45^{\circ}

3

60^{\circ}

4

\tan^{- 1} (\frac{36}{67})

Explanation:

B Radius of circular path
$r = 134 ft$
$= 134 \times 0.3045 m = 40.84$ meter
Velocity $V = 72 km / hr$
$= \frac{72 \times 1000}{3600} = 20 m / \sec$
The angle of inclination
$∵ \tan θ = \frac{V^{2}}{rg}$
$\tan θ = \frac{(20)^{2}}{40.84 \times 9.8} = \frac{400}{403.08} = 0.9996$
$\tan θ ≃ 1$
$θ = \tan^{- 1} (1) = 45^{\circ}$
Angle of inclination $(θ) = 45^{\circ}$

Tripura-2020

Motion in One Dimensions

141206 The instantaneous velocity of a particle moving in a straight line is given as $v = α t + β t^{2}$ , where $α$ and $β$ are constants. The distance travelled by the particle between $1 s$ and $2 s$ is

1

3 α + 7 β

2

\frac{3}{2} α + \frac{7}{3} β

3

\frac{α}{2} + \frac{β}{3}

4

\frac{3}{2} α + \frac{7}{2} β

Explanation:

B The instantaneous velocity of a particle is given as -
$v = α t + β t^{2}$
$∵ v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t + β t^{2}$
$d x = (α t + β t^{2}) d t$
Integrating the above equation from, $t = 1$ to 2
$\int_{1}^{2} dx = \int_{1}^{2} [α t] dt + \int_{1}^{2} β t^{2} dt$
$[x]_{1}^{2} = \frac{[α t]_{1}^{2}}{2} + \frac{{[β t^{3}]}_{1}^{2}}{3}$
$[x]_{1}^{2} = \frac{α}{2} [2^{2} - 1^{1}] + \frac{β}{3} [2^{3} - 1^{3}]$
$[x]_{1}^{2} = \frac{α}{2} [4 - 1] + \frac{β}{3} [8 - 1]$
$Δ x = (x_{2} - x_{1}) = \frac{3 α}{2} + \frac{7 β}{3}$

JEE Main-25.07.2021

Motion in One Dimensions

141207 A time dependent force is acting on a body of which the velocity of body can be expressed as $α t^{3 / 2}$ . The displacement of the body between time $t = 2$ and $5 \sec$ is about.

1

20 α

2

22 α

3

6 α

4 0

Explanation:

A Velocity of the body is given as -
$v = α t^{3 / 2}$
To find out the displacement between time $(t) = 2$ second to 5 second
$v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t^{3 / 2}$
$dx = α t^{3 / 2} dt$
Integrating the above equation from $t = 2$ to $5 s -$
$\int_{2}^{5} dx = \int_{2}^{5} α t^{3 / 2} dt$
$[x_{5} - x_{2}] = α {[\frac{t^{5 / 2}}{5 / 2}]}_{2}^{5}$
$Δ x = α \times \frac{2}{5} [5^{2.5} - 2^{2.5}]$
$Δ x = α \times \frac{2}{5} [55.90 - 5.65]$
$Δ x = \frac{2 α}{5} [50.24]$
$Δ x = 20.09 α \approx 20 α$
$Δ x = 20 α$

Assam CEE-2021

Motion in One Dimensions

141208 An object is moving along $x$ -axis. If its position $x (t)$ (in meter) at any time $t$ (in $s$ ) is given by the following graph, then the object is at rest about
original image

1

t = - 1 s

2

t = 0 s

3

t = 1 s

4

t = 2 s

Explanation:

C An object is moving along $x$ -axis. If its position $x (t)$ in meter at any time $+ ($ in sec) is given by the following graph. Then the object is at rest about $t = 1 \sec$
Because at $t = 1 \sec$ position of object is not changing.
original image

TS EAMCET 28.09.2020

Motion in One Dimensions

141209 A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of $72 km / hr$ . The angle of inclination of the bike with horizontal plant will be

1

30^{\circ}

2

45^{\circ}

3

60^{\circ}

4

\tan^{- 1} (\frac{36}{67})

Explanation:

B Radius of circular path
$r = 134 ft$
$= 134 \times 0.3045 m = 40.84$ meter
Velocity $V = 72 km / hr$
$= \frac{72 \times 1000}{3600} = 20 m / \sec$
The angle of inclination
$∵ \tan θ = \frac{V^{2}}{rg}$
$\tan θ = \frac{(20)^{2}}{40.84 \times 9.8} = \frac{400}{403.08} = 0.9996$
$\tan θ ≃ 1$
$θ = \tan^{- 1} (1) = 45^{\circ}$
Angle of inclination $(θ) = 45^{\circ}$

Tripura-2020

Motion in One Dimensions

141206 The instantaneous velocity of a particle moving in a straight line is given as $v = α t + β t^{2}$ , where $α$ and $β$ are constants. The distance travelled by the particle between $1 s$ and $2 s$ is

1

3 α + 7 β

2

\frac{3}{2} α + \frac{7}{3} β

3

\frac{α}{2} + \frac{β}{3}

4

\frac{3}{2} α + \frac{7}{2} β

Explanation:

B The instantaneous velocity of a particle is given as -
$v = α t + β t^{2}$
$∵ v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t + β t^{2}$
$d x = (α t + β t^{2}) d t$
Integrating the above equation from, $t = 1$ to 2
$\int_{1}^{2} dx = \int_{1}^{2} [α t] dt + \int_{1}^{2} β t^{2} dt$
$[x]_{1}^{2} = \frac{[α t]_{1}^{2}}{2} + \frac{{[β t^{3}]}_{1}^{2}}{3}$
$[x]_{1}^{2} = \frac{α}{2} [2^{2} - 1^{1}] + \frac{β}{3} [2^{3} - 1^{3}]$
$[x]_{1}^{2} = \frac{α}{2} [4 - 1] + \frac{β}{3} [8 - 1]$
$Δ x = (x_{2} - x_{1}) = \frac{3 α}{2} + \frac{7 β}{3}$

JEE Main-25.07.2021

Motion in One Dimensions

141207 A time dependent force is acting on a body of which the velocity of body can be expressed as $α t^{3 / 2}$ . The displacement of the body between time $t = 2$ and $5 \sec$ is about.

1

20 α

2

22 α

3

6 α

4 0

Explanation:

A Velocity of the body is given as -
$v = α t^{3 / 2}$
To find out the displacement between time $(t) = 2$ second to 5 second
$v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t^{3 / 2}$
$dx = α t^{3 / 2} dt$
Integrating the above equation from $t = 2$ to $5 s -$
$\int_{2}^{5} dx = \int_{2}^{5} α t^{3 / 2} dt$
$[x_{5} - x_{2}] = α {[\frac{t^{5 / 2}}{5 / 2}]}_{2}^{5}$
$Δ x = α \times \frac{2}{5} [5^{2.5} - 2^{2.5}]$
$Δ x = α \times \frac{2}{5} [55.90 - 5.65]$
$Δ x = \frac{2 α}{5} [50.24]$
$Δ x = 20.09 α \approx 20 α$
$Δ x = 20 α$

Assam CEE-2021

Motion in One Dimensions

141208 An object is moving along $x$ -axis. If its position $x (t)$ (in meter) at any time $t$ (in $s$ ) is given by the following graph, then the object is at rest about
original image

1

t = - 1 s

2

t = 0 s

3

t = 1 s

4

t = 2 s

Explanation:

C An object is moving along $x$ -axis. If its position $x (t)$ in meter at any time $+ ($ in sec) is given by the following graph. Then the object is at rest about $t = 1 \sec$
Because at $t = 1 \sec$ position of object is not changing.
original image

TS EAMCET 28.09.2020

Motion in One Dimensions

141209 A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of $72 km / hr$ . The angle of inclination of the bike with horizontal plant will be

1

30^{\circ}

2

45^{\circ}

3

60^{\circ}

4

\tan^{- 1} (\frac{36}{67})

Explanation:

B Radius of circular path
$r = 134 ft$
$= 134 \times 0.3045 m = 40.84$ meter
Velocity $V = 72 km / hr$
$= \frac{72 \times 1000}{3600} = 20 m / \sec$
The angle of inclination
$∵ \tan θ = \frac{V^{2}}{rg}$
$\tan θ = \frac{(20)^{2}}{40.84 \times 9.8} = \frac{400}{403.08} = 0.9996$
$\tan θ ≃ 1$
$θ = \tan^{- 1} (1) = 45^{\circ}$
Angle of inclination $(θ) = 45^{\circ}$

Tripura-2020

Motion in One Dimensions

141206 The instantaneous velocity of a particle moving in a straight line is given as $v = α t + β t^{2}$ , where $α$ and $β$ are constants. The distance travelled by the particle between $1 s$ and $2 s$ is

1

3 α + 7 β

2

\frac{3}{2} α + \frac{7}{3} β

3

\frac{α}{2} + \frac{β}{3}

4

\frac{3}{2} α + \frac{7}{2} β

Explanation:

B The instantaneous velocity of a particle is given as -
$v = α t + β t^{2}$
$∵ v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t + β t^{2}$
$d x = (α t + β t^{2}) d t$
Integrating the above equation from, $t = 1$ to 2
$\int_{1}^{2} dx = \int_{1}^{2} [α t] dt + \int_{1}^{2} β t^{2} dt$
$[x]_{1}^{2} = \frac{[α t]_{1}^{2}}{2} + \frac{{[β t^{3}]}_{1}^{2}}{3}$
$[x]_{1}^{2} = \frac{α}{2} [2^{2} - 1^{1}] + \frac{β}{3} [2^{3} - 1^{3}]$
$[x]_{1}^{2} = \frac{α}{2} [4 - 1] + \frac{β}{3} [8 - 1]$
$Δ x = (x_{2} - x_{1}) = \frac{3 α}{2} + \frac{7 β}{3}$

JEE Main-25.07.2021

Motion in One Dimensions

141207 A time dependent force is acting on a body of which the velocity of body can be expressed as $α t^{3 / 2}$ . The displacement of the body between time $t = 2$ and $5 \sec$ is about.

1

20 α

2

22 α

3

6 α

4 0

Explanation:

A Velocity of the body is given as -
$v = α t^{3 / 2}$
To find out the displacement between time $(t) = 2$ second to 5 second
$v = \frac{dx}{dt}$
$∴ \frac{dx}{dt} = α t^{3 / 2}$
$dx = α t^{3 / 2} dt$
Integrating the above equation from $t = 2$ to $5 s -$
$\int_{2}^{5} dx = \int_{2}^{5} α t^{3 / 2} dt$
$[x_{5} - x_{2}] = α {[\frac{t^{5 / 2}}{5 / 2}]}_{2}^{5}$
$Δ x = α \times \frac{2}{5} [5^{2.5} - 2^{2.5}]$
$Δ x = α \times \frac{2}{5} [55.90 - 5.65]$
$Δ x = \frac{2 α}{5} [50.24]$
$Δ x = 20.09 α \approx 20 α$
$Δ x = 20 α$

Assam CEE-2021

Motion in One Dimensions

141208 An object is moving along $x$ -axis. If its position $x (t)$ (in meter) at any time $t$ (in $s$ ) is given by the following graph, then the object is at rest about
original image

1

t = - 1 s

2

t = 0 s

3

t = 1 s

4

t = 2 s

Explanation:

C An object is moving along $x$ -axis. If its position $x (t)$ in meter at any time $+ ($ in sec) is given by the following graph. Then the object is at rest about $t = 1 \sec$
Because at $t = 1 \sec$ position of object is not changing.
original image

TS EAMCET 28.09.2020

Motion in One Dimensions

141209 A person on bike is moving along a horizontal circular path of radius 134 feet a constant speed of $72 km / hr$ . The angle of inclination of the bike with horizontal plant will be

1

30^{\circ}

2

45^{\circ}

3

60^{\circ}

4

\tan^{- 1} (\frac{36}{67})

Explanation:

B Radius of circular path
$r = 134 ft$
$= 134 \times 0.3045 m = 40.84$ meter
Velocity $V = 72 km / hr$
$= \frac{72 \times 1000}{3600} = 20 m / \sec$
The angle of inclination
$∵ \tan θ = \frac{V^{2}}{rg}$
$\tan θ = \frac{(20)^{2}}{40.84 \times 9.8} = \frac{400}{403.08} = 0.9996$
$\tan θ ≃ 1$
$θ = \tan^{- 1} (1) = 45^{\circ}$
Angle of inclination $(θ) = 45^{\circ}$

Tripura-2020