QBManager

Motion in One Dimensions

141494 Displacement $x$ versus $t^{2}$ graph is shown for a particle. The acceleration of the particle is
$x (m)$
original image

1

4 m / s^{2}

2

8 m / s^{2}

3 zero

4

2 m / s^{2}

Explanation:

A original image
By given graph
$x = slope (t^{2})$
$x = \frac{2 - 0}{1 - 0} \times t^{2}$
$x = 2 t^{2}$
Differentiating $x$ w.r.t ' $t$ ' we get velocity
$\frac{dx}{dt} = 4 t$
Again differential $\frac{dx}{dt}$ w.r.t ' $t$ ' we get acceleration
$\frac{d^{2} x}{{dt}^{2}} = 4$
Which is the value of acceleration
$i.e. a = 4 m / s^{2}$

UPSEE - 2017

Motion in One Dimensions

141497 A packet of weight ' $W$ ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is

1

W

2

2 W

3

3 W

4

4 W

Explanation:

C Given,
packet weight $= w$ , Retardation $(a) = 2 g$
original image
$w = mg$
From figure
$F - w = ma$
$F - w = \frac{w}{g} \times 2 g \Rightarrow F - w = 2 w$
$F = 3 w$

AP EAMCET-25.04.2017

Motion in One Dimensions

141498 The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between $t = 4 s$ and $t$ $= 6 s$ is
original image

1

10.5 {ms}^{- 1}

2

12.5 {ms}^{- 1}

3

7.5 {ms}^{- 1}

4

9.5 {ms}^{- 1}

Explanation:

B Given,
time interval, $t = t_{2} - t_{1} = 2 \sec$
$t_{1} = 4 \sec, t_{2} = 6 \sec$
slop of v-t graph,
$[v = velocity, t = time]$
We know that,
Acceleration, $a = \tan θ = \frac{15}{6} = \frac{5}{2} {ms}^{- 2}$
initial $u = 0$
velocity at $t_{1} = 4 \sec$
then, $v = u + a t$
$v = 0 + \frac{5}{2} \times 4$
$v_{1} = 5 \times 2 = 10 m / \sec (given)$
So, average
$\frac{v_{1} + v_{2}}{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5 = 12.5 m / \sec$

AP EAMCET-25.04.2017

Motion in One Dimensions

141499 The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
original image

1

\frac{20}{7} {ms}^{- 1}

2

\frac{18}{7} {ms}^{- 1}

3

\frac{36}{7} {ms}^{- 1}

4

\frac{12}{7} {ms}^{- 1}

Explanation:

B $V_{Avg} = \frac{Total distance}{total time}$
Total distance $= S$
$S_{1} = \frac{1}{2} \times 2 \times 6 = 6 m$
$S_{2} = 6 \times 2 = 12$
$S = S_{1} + S_{2}$
$= 6 + 12 = 18 m$
$V_{Avg} = \frac{18}{7} m / \sec .$

AP EAMCET-28.04.2017

Motion in One Dimensions

141494 Displacement $x$ versus $t^{2}$ graph is shown for a particle. The acceleration of the particle is
$x (m)$
original image

1

4 m / s^{2}

2

8 m / s^{2}

3 zero

4

2 m / s^{2}

Explanation:

A original image
By given graph
$x = slope (t^{2})$
$x = \frac{2 - 0}{1 - 0} \times t^{2}$
$x = 2 t^{2}$
Differentiating $x$ w.r.t ' $t$ ' we get velocity
$\frac{dx}{dt} = 4 t$
Again differential $\frac{dx}{dt}$ w.r.t ' $t$ ' we get acceleration
$\frac{d^{2} x}{{dt}^{2}} = 4$
Which is the value of acceleration
$i.e. a = 4 m / s^{2}$

UPSEE - 2017

Motion in One Dimensions

141497 A packet of weight ' $W$ ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is

1

W

2

2 W

3

3 W

4

4 W

Explanation:

C Given,
packet weight $= w$ , Retardation $(a) = 2 g$
original image
$w = mg$
From figure
$F - w = ma$
$F - w = \frac{w}{g} \times 2 g \Rightarrow F - w = 2 w$
$F = 3 w$

AP EAMCET-25.04.2017

Motion in One Dimensions

141498 The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between $t = 4 s$ and $t$ $= 6 s$ is
original image

1

10.5 {ms}^{- 1}

2

12.5 {ms}^{- 1}

3

7.5 {ms}^{- 1}

4

9.5 {ms}^{- 1}

Explanation:

B Given,
time interval, $t = t_{2} - t_{1} = 2 \sec$
$t_{1} = 4 \sec, t_{2} = 6 \sec$
slop of v-t graph,
$[v = velocity, t = time]$
We know that,
Acceleration, $a = \tan θ = \frac{15}{6} = \frac{5}{2} {ms}^{- 2}$
initial $u = 0$
velocity at $t_{1} = 4 \sec$
then, $v = u + a t$
$v = 0 + \frac{5}{2} \times 4$
$v_{1} = 5 \times 2 = 10 m / \sec (given)$
So, average
$\frac{v_{1} + v_{2}}{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5 = 12.5 m / \sec$

AP EAMCET-25.04.2017

Motion in One Dimensions

141499 The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
original image

1

\frac{20}{7} {ms}^{- 1}

2

\frac{18}{7} {ms}^{- 1}

3

\frac{36}{7} {ms}^{- 1}

4

\frac{12}{7} {ms}^{- 1}

Explanation:

B $V_{Avg} = \frac{Total distance}{total time}$
Total distance $= S$
$S_{1} = \frac{1}{2} \times 2 \times 6 = 6 m$
$S_{2} = 6 \times 2 = 12$
$S = S_{1} + S_{2}$
$= 6 + 12 = 18 m$
$V_{Avg} = \frac{18}{7} m / \sec .$

AP EAMCET-28.04.2017

Motion in One Dimensions

141494 Displacement $x$ versus $t^{2}$ graph is shown for a particle. The acceleration of the particle is
$x (m)$
original image

1

4 m / s^{2}

2

8 m / s^{2}

3 zero

4

2 m / s^{2}

Explanation:

A original image
By given graph
$x = slope (t^{2})$
$x = \frac{2 - 0}{1 - 0} \times t^{2}$
$x = 2 t^{2}$
Differentiating $x$ w.r.t ' $t$ ' we get velocity
$\frac{dx}{dt} = 4 t$
Again differential $\frac{dx}{dt}$ w.r.t ' $t$ ' we get acceleration
$\frac{d^{2} x}{{dt}^{2}} = 4$
Which is the value of acceleration
$i.e. a = 4 m / s^{2}$

UPSEE - 2017

Motion in One Dimensions

141497 A packet of weight ' $W$ ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is

1

W

2

2 W

3

3 W

4

4 W

Explanation:

C Given,
packet weight $= w$ , Retardation $(a) = 2 g$
original image
$w = mg$
From figure
$F - w = ma$
$F - w = \frac{w}{g} \times 2 g \Rightarrow F - w = 2 w$
$F = 3 w$

AP EAMCET-25.04.2017

Motion in One Dimensions

141498 The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between $t = 4 s$ and $t$ $= 6 s$ is
original image

1

10.5 {ms}^{- 1}

2

12.5 {ms}^{- 1}

3

7.5 {ms}^{- 1}

4

9.5 {ms}^{- 1}

Explanation:

B Given,
time interval, $t = t_{2} - t_{1} = 2 \sec$
$t_{1} = 4 \sec, t_{2} = 6 \sec$
slop of v-t graph,
$[v = velocity, t = time]$
We know that,
Acceleration, $a = \tan θ = \frac{15}{6} = \frac{5}{2} {ms}^{- 2}$
initial $u = 0$
velocity at $t_{1} = 4 \sec$
then, $v = u + a t$
$v = 0 + \frac{5}{2} \times 4$
$v_{1} = 5 \times 2 = 10 m / \sec (given)$
So, average
$\frac{v_{1} + v_{2}}{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5 = 12.5 m / \sec$

AP EAMCET-25.04.2017

Motion in One Dimensions

141499 The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
original image

1

\frac{20}{7} {ms}^{- 1}

2

\frac{18}{7} {ms}^{- 1}

3

\frac{36}{7} {ms}^{- 1}

4

\frac{12}{7} {ms}^{- 1}

Explanation:

B $V_{Avg} = \frac{Total distance}{total time}$
Total distance $= S$
$S_{1} = \frac{1}{2} \times 2 \times 6 = 6 m$
$S_{2} = 6 \times 2 = 12$
$S = S_{1} + S_{2}$
$= 6 + 12 = 18 m$
$V_{Avg} = \frac{18}{7} m / \sec .$

AP EAMCET-28.04.2017

Motion in One Dimensions

141494 Displacement $x$ versus $t^{2}$ graph is shown for a particle. The acceleration of the particle is
$x (m)$
original image

1

4 m / s^{2}

2

8 m / s^{2}

3 zero

4

2 m / s^{2}

Explanation:

A original image
By given graph
$x = slope (t^{2})$
$x = \frac{2 - 0}{1 - 0} \times t^{2}$
$x = 2 t^{2}$
Differentiating $x$ w.r.t ' $t$ ' we get velocity
$\frac{dx}{dt} = 4 t$
Again differential $\frac{dx}{dt}$ w.r.t ' $t$ ' we get acceleration
$\frac{d^{2} x}{{dt}^{2}} = 4$
Which is the value of acceleration
$i.e. a = 4 m / s^{2}$

UPSEE - 2017

Motion in One Dimensions

141497 A packet of weight ' $W$ ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is

1

W

2

2 W

3

3 W

4

4 W

Explanation:

C Given,
packet weight $= w$ , Retardation $(a) = 2 g$
original image
$w = mg$
From figure
$F - w = ma$
$F - w = \frac{w}{g} \times 2 g \Rightarrow F - w = 2 w$
$F = 3 w$

AP EAMCET-25.04.2017

Motion in One Dimensions

141498 The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between $t = 4 s$ and $t$ $= 6 s$ is
original image

1

10.5 {ms}^{- 1}

2

12.5 {ms}^{- 1}

3

7.5 {ms}^{- 1}

4

9.5 {ms}^{- 1}

Explanation:

B Given,
time interval, $t = t_{2} - t_{1} = 2 \sec$
$t_{1} = 4 \sec, t_{2} = 6 \sec$
slop of v-t graph,
$[v = velocity, t = time]$
We know that,
Acceleration, $a = \tan θ = \frac{15}{6} = \frac{5}{2} {ms}^{- 2}$
initial $u = 0$
velocity at $t_{1} = 4 \sec$
then, $v = u + a t$
$v = 0 + \frac{5}{2} \times 4$
$v_{1} = 5 \times 2 = 10 m / \sec (given)$
So, average
$\frac{v_{1} + v_{2}}{2} = \frac{10 + 15}{2} = \frac{25}{2} = 12.5 = 12.5 m / \sec$

AP EAMCET-25.04.2017

Motion in One Dimensions

141499 The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
original image

1

\frac{20}{7} {ms}^{- 1}

2

\frac{18}{7} {ms}^{- 1}

3

\frac{36}{7} {ms}^{- 1}

4

\frac{12}{7} {ms}^{- 1}

Explanation:

B $V_{Avg} = \frac{Total distance}{total time}$
Total distance $= S$
$S_{1} = \frac{1}{2} \times 2 \times 6 = 6 m$
$S_{2} = 6 \times 2 = 12$
$S = S_{1} + S_{2}$
$= 6 + 12 = 18 m$
$V_{Avg} = \frac{18}{7} m / \sec .$

AP EAMCET-28.04.2017

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