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Motion in One Dimensions

141210 When a small object is thrown vertically upward its velocity versus time graph is represented as follows
original image
The maximum height it can reach is

1

\frac{a^{2}}{b}

2

\frac{b^{2}}{a}

3

2 ab

4

ab

Explanation:

C original image
$\frac{ds}{dt} = v$
$ds = vdt$
$ds =$ area under the graph
${ds}_{1} = \frac{1}{2} \times b \times 2 a = ab$
${ds}_{2} = \frac{1}{2} \times b \times - 2 a = - ab$
displacement $= {ds}_{1} + {ds}_{2}$
$= a b - a b = 0$
(Height) distance $= ({ds}_{1}) + | {ds}_{2} |$
$= a b + | - a b |$
$= a b + a b = 2 a b$

Tripura-2020

Motion in One Dimensions

141212 The velocity (v) of a body moving along the positive $x$ direction varies with displacement ( $x$ ) from the origin as $v = k \sqrt{x}$ , where $k$ is a constant. Then which of the following $x - t$ graph is correct.

1

2

3

4

Explanation:

C Given velocity displacement function
$v = k \sqrt{x} \Rightarrow \frac{dx}{dt} = k \sqrt{x}$
$\int \frac{dx}{\sqrt{x}} = \int kdt$
$2 \sqrt{x} = kt$
$t = \frac{2 \sqrt{x}}{k}$
$t^{2} = \frac{4 x}{k^{2}}$
i.e. $y^{2} = 4 a x \dots \dots$ which is equation of parabola.

Assam CEE-2020

Motion in One Dimensions

141213 A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 km h^{- 1}$ next $1 / 3$ of the distance with velocity $20 km h^{- 1}$ and the last $1 / 3$ of the distance with velocity $60 km h^{- 1}$ . The average velocity of the car in the whole journey is

1

18 km h^{- 1}

2

30 km h^{- 1}

3

36 km h^{- 1}

4

90 km h^{- 1}

Explanation:

A original image
$\overset{\leftarrow}{\leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶∣}$
$∵ Average velocity = \frac{total distance}{total time}$
Average velocity $= \frac{D}{(\frac{D / 3}{10} + \frac{D / 3}{20} + \frac{D / 3}{60})}$
$= \frac{D}{\frac{D}{3} \times \frac{1}{10} (1 + \frac{1}{2} + \frac{1}{6})}$
$= \frac{30 \times 6}{6 + 3 + 1} = \frac{30 \times 6}{10} = 18 km h^{- 1}$

Assam CEE-2020

Motion in One Dimensions

141214 The v-t graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333 s$ . The total distance covered by the body in $6 s$ is
$v$ (in $m / s$ ) 4
original image

1

\frac{37}{3} m

2

12 m

3

11 m

4

\frac{49}{4} m

Explanation:

A original image
Distance cover by the body in $6 \sec$ is-
$=$ Area of $△ AOE +$ Area of rectangle ABFE + Area of $△ BSF +$ Area of $△ SGC +$ Area of $△ GCD$
$= \frac{1}{2} (2 \times 4) + (1 \times 4) + \frac{1}{2} \times 4 \times (\frac{4}{3}) + \frac{1}{2} \times 2 \times (\frac{2}{3}) + \frac{1}{2} \times 2 \times 1$
$= 4 + 4 + \frac{8}{3} + \frac{2}{3} + 1 = \frac{37}{3} m$

JEE Main-05.09.2020

Motion in One Dimensions

141210 When a small object is thrown vertically upward its velocity versus time graph is represented as follows
original image
The maximum height it can reach is

1

\frac{a^{2}}{b}

2

\frac{b^{2}}{a}

3

2 ab

4

ab

Explanation:

C original image
$\frac{ds}{dt} = v$
$ds = vdt$
$ds =$ area under the graph
${ds}_{1} = \frac{1}{2} \times b \times 2 a = ab$
${ds}_{2} = \frac{1}{2} \times b \times - 2 a = - ab$
displacement $= {ds}_{1} + {ds}_{2}$
$= a b - a b = 0$
(Height) distance $= ({ds}_{1}) + | {ds}_{2} |$
$= a b + | - a b |$
$= a b + a b = 2 a b$

Tripura-2020

Motion in One Dimensions

141212 The velocity (v) of a body moving along the positive $x$ direction varies with displacement ( $x$ ) from the origin as $v = k \sqrt{x}$ , where $k$ is a constant. Then which of the following $x - t$ graph is correct.

1

2

3

4

Explanation:

C Given velocity displacement function
$v = k \sqrt{x} \Rightarrow \frac{dx}{dt} = k \sqrt{x}$
$\int \frac{dx}{\sqrt{x}} = \int kdt$
$2 \sqrt{x} = kt$
$t = \frac{2 \sqrt{x}}{k}$
$t^{2} = \frac{4 x}{k^{2}}$
i.e. $y^{2} = 4 a x \dots \dots$ which is equation of parabola.

Assam CEE-2020

Motion in One Dimensions

141213 A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 km h^{- 1}$ next $1 / 3$ of the distance with velocity $20 km h^{- 1}$ and the last $1 / 3$ of the distance with velocity $60 km h^{- 1}$ . The average velocity of the car in the whole journey is

1

18 km h^{- 1}

2

30 km h^{- 1}

3

36 km h^{- 1}

4

90 km h^{- 1}

Explanation:

A original image
$\overset{\leftarrow}{\leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶∣}$
$∵ Average velocity = \frac{total distance}{total time}$
Average velocity $= \frac{D}{(\frac{D / 3}{10} + \frac{D / 3}{20} + \frac{D / 3}{60})}$
$= \frac{D}{\frac{D}{3} \times \frac{1}{10} (1 + \frac{1}{2} + \frac{1}{6})}$
$= \frac{30 \times 6}{6 + 3 + 1} = \frac{30 \times 6}{10} = 18 km h^{- 1}$

Assam CEE-2020

Motion in One Dimensions

141214 The v-t graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333 s$ . The total distance covered by the body in $6 s$ is
$v$ (in $m / s$ ) 4
original image

1

\frac{37}{3} m

2

12 m

3

11 m

4

\frac{49}{4} m

Explanation:

A original image
Distance cover by the body in $6 \sec$ is-
$=$ Area of $△ AOE +$ Area of rectangle ABFE + Area of $△ BSF +$ Area of $△ SGC +$ Area of $△ GCD$
$= \frac{1}{2} (2 \times 4) + (1 \times 4) + \frac{1}{2} \times 4 \times (\frac{4}{3}) + \frac{1}{2} \times 2 \times (\frac{2}{3}) + \frac{1}{2} \times 2 \times 1$
$= 4 + 4 + \frac{8}{3} + \frac{2}{3} + 1 = \frac{37}{3} m$

JEE Main-05.09.2020

Motion in One Dimensions

141210 When a small object is thrown vertically upward its velocity versus time graph is represented as follows
original image
The maximum height it can reach is

1

\frac{a^{2}}{b}

2

\frac{b^{2}}{a}

3

2 ab

4

ab

Explanation:

C original image
$\frac{ds}{dt} = v$
$ds = vdt$
$ds =$ area under the graph
${ds}_{1} = \frac{1}{2} \times b \times 2 a = ab$
${ds}_{2} = \frac{1}{2} \times b \times - 2 a = - ab$
displacement $= {ds}_{1} + {ds}_{2}$
$= a b - a b = 0$
(Height) distance $= ({ds}_{1}) + | {ds}_{2} |$
$= a b + | - a b |$
$= a b + a b = 2 a b$

Tripura-2020

Motion in One Dimensions

141212 The velocity (v) of a body moving along the positive $x$ direction varies with displacement ( $x$ ) from the origin as $v = k \sqrt{x}$ , where $k$ is a constant. Then which of the following $x - t$ graph is correct.

1

2

3

4

Explanation:

C Given velocity displacement function
$v = k \sqrt{x} \Rightarrow \frac{dx}{dt} = k \sqrt{x}$
$\int \frac{dx}{\sqrt{x}} = \int kdt$
$2 \sqrt{x} = kt$
$t = \frac{2 \sqrt{x}}{k}$
$t^{2} = \frac{4 x}{k^{2}}$
i.e. $y^{2} = 4 a x \dots \dots$ which is equation of parabola.

Assam CEE-2020

Motion in One Dimensions

141213 A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 km h^{- 1}$ next $1 / 3$ of the distance with velocity $20 km h^{- 1}$ and the last $1 / 3$ of the distance with velocity $60 km h^{- 1}$ . The average velocity of the car in the whole journey is

1

18 km h^{- 1}

2

30 km h^{- 1}

3

36 km h^{- 1}

4

90 km h^{- 1}

Explanation:

A original image
$\overset{\leftarrow}{\leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶∣}$
$∵ Average velocity = \frac{total distance}{total time}$
Average velocity $= \frac{D}{(\frac{D / 3}{10} + \frac{D / 3}{20} + \frac{D / 3}{60})}$
$= \frac{D}{\frac{D}{3} \times \frac{1}{10} (1 + \frac{1}{2} + \frac{1}{6})}$
$= \frac{30 \times 6}{6 + 3 + 1} = \frac{30 \times 6}{10} = 18 km h^{- 1}$

Assam CEE-2020

Motion in One Dimensions

141214 The v-t graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333 s$ . The total distance covered by the body in $6 s$ is
$v$ (in $m / s$ ) 4
original image

1

\frac{37}{3} m

2

12 m

3

11 m

4

\frac{49}{4} m

Explanation:

A original image
Distance cover by the body in $6 \sec$ is-
$=$ Area of $△ AOE +$ Area of rectangle ABFE + Area of $△ BSF +$ Area of $△ SGC +$ Area of $△ GCD$
$= \frac{1}{2} (2 \times 4) + (1 \times 4) + \frac{1}{2} \times 4 \times (\frac{4}{3}) + \frac{1}{2} \times 2 \times (\frac{2}{3}) + \frac{1}{2} \times 2 \times 1$
$= 4 + 4 + \frac{8}{3} + \frac{2}{3} + 1 = \frac{37}{3} m$

JEE Main-05.09.2020

Motion in One Dimensions

141210 When a small object is thrown vertically upward its velocity versus time graph is represented as follows
original image
The maximum height it can reach is

1

\frac{a^{2}}{b}

2

\frac{b^{2}}{a}

3

2 ab

4

ab

Explanation:

C original image
$\frac{ds}{dt} = v$
$ds = vdt$
$ds =$ area under the graph
${ds}_{1} = \frac{1}{2} \times b \times 2 a = ab$
${ds}_{2} = \frac{1}{2} \times b \times - 2 a = - ab$
displacement $= {ds}_{1} + {ds}_{2}$
$= a b - a b = 0$
(Height) distance $= ({ds}_{1}) + | {ds}_{2} |$
$= a b + | - a b |$
$= a b + a b = 2 a b$

Tripura-2020

Motion in One Dimensions

141212 The velocity (v) of a body moving along the positive $x$ direction varies with displacement ( $x$ ) from the origin as $v = k \sqrt{x}$ , where $k$ is a constant. Then which of the following $x - t$ graph is correct.

1

2

3

4

Explanation:

C Given velocity displacement function
$v = k \sqrt{x} \Rightarrow \frac{dx}{dt} = k \sqrt{x}$
$\int \frac{dx}{\sqrt{x}} = \int kdt$
$2 \sqrt{x} = kt$
$t = \frac{2 \sqrt{x}}{k}$
$t^{2} = \frac{4 x}{k^{2}}$
i.e. $y^{2} = 4 a x \dots \dots$ which is equation of parabola.

Assam CEE-2020

Motion in One Dimensions

141213 A car travels $1 / 3$ of the distance on a straight road with a velocity of $10 km h^{- 1}$ next $1 / 3$ of the distance with velocity $20 km h^{- 1}$ and the last $1 / 3$ of the distance with velocity $60 km h^{- 1}$ . The average velocity of the car in the whole journey is

1

18 km h^{- 1}

2

30 km h^{- 1}

3

36 km h^{- 1}

4

90 km h^{- 1}

Explanation:

A original image
$\overset{\leftarrow}{\leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶ | \leftarrow D / 3 ⟶∣}$
$∵ Average velocity = \frac{total distance}{total time}$
Average velocity $= \frac{D}{(\frac{D / 3}{10} + \frac{D / 3}{20} + \frac{D / 3}{60})}$
$= \frac{D}{\frac{D}{3} \times \frac{1}{10} (1 + \frac{1}{2} + \frac{1}{6})}$
$= \frac{30 \times 6}{6 + 3 + 1} = \frac{30 \times 6}{10} = 18 km h^{- 1}$

Assam CEE-2020

Motion in One Dimensions

141214 The v-t graph of a body in a straight line motion is shown in the figure. The point $S$ is at $4.333 s$ . The total distance covered by the body in $6 s$ is
$v$ (in $m / s$ ) 4
original image

1

\frac{37}{3} m

2

12 m

3

11 m

4

\frac{49}{4} m

Explanation:

A original image
Distance cover by the body in $6 \sec$ is-
$=$ Area of $△ AOE +$ Area of rectangle ABFE + Area of $△ BSF +$ Area of $△ SGC +$ Area of $△ GCD$
$= \frac{1}{2} (2 \times 4) + (1 \times 4) + \frac{1}{2} \times 4 \times (\frac{4}{3}) + \frac{1}{2} \times 2 \times (\frac{2}{3}) + \frac{1}{2} \times 2 \times 1$
$= 4 + 4 + \frac{8}{3} + \frac{2}{3} + 1 = \frac{37}{3} m$

JEE Main-05.09.2020

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