QBManager

Motion in One Dimensions

141255 The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
original image

1 The total distance travelled by the particle is zero

2 The total displacement of the particle is zero

3 The maximum acceleration of the particle is

\frac{1}{2} {ms}^{- 2}

4 The total distance travelled by the particle at the end of

10 s

is

100 m

5 At the fifth second, the acceleration of the particle is

2 {ms}^{- 2}

Explanation:

B $v = \frac{ds}{dt}$
$\int ds = \int vdt$
$Δ s =$ Area under v-t curve
$Δ s = d_{1} + d_{2}$
$Δ s = \frac{1}{2} \times 10 \times 10 - \frac{1}{2} \times 10 \times 10$
$∴ Δ s = 0$

Kerala CEE -2018

Motion in One Dimensions

141257 A car starts from rest with a constant acceleration of $5 {ms}^{- 2}$ . At the same time a bus travelling with uniform velocity of $50 {ms}^{- 1}$ overtakes and passes the car. The distance through which the car again overtakes the bus is:

1

250 m

2

500 m

3

750 m

4

1000 m

Explanation:

D Given, $s = 50 t, u = 0, a = 5 m / s^{2}$
Distance covered by bus will be equal to distance covered by car.
$∵ s = ut + \frac{1}{2} {at}^{2}$
$50 t = 0 \times t + \frac{1}{2} 5 t^{2}$
$t = 20 \sec$
Distance $=$ speed $\times$ time $= 50 \times 20 = 1000 m$

Assam CEE-2017

Motion in One Dimensions

141258 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

1

2

3

4

Explanation:

B $v = u - gt = u - 10 t (∵ g = 10 m / s^{2})$ $v - t$ graph
original image
$y = u - 10 x$ (straight line)
$y = c + mx$
compare-
Slope $(m) = - 10 (- ve)$
$c = u (+ ve)$

JEE Main 2017

Motion in One Dimensions

141259 The displacement of a particle moving along $x$ axis versus time is given in the figure below: The average velocity $(V_{avg})$ of the particle in the first four seconds and the velocity $V$ of it at $t = 4 s$ are
original image

1

V_{arg} = - 1 m / s, V = - 1 m / s

2

V_{avg} = 1 m / s, V = 1 m / s

3

V_{avg} = 1 m / s, V = 0 m / s

4

V_{avg} = - 1 m / s, V = 0 m / s

Explanation:

D Let us consider
As $t = 0$ , particle is at a position $5 m$ , and at $t = 4 s$ particle is at a position $1 m$ .
Displacement of particle $(s) = 1 - 5 = - 4 m$ time taken $(t) = 4 s$
original image
Average velocity $(V_{avg}) = \frac{s}{t} = \frac{- 4}{4} = - 1 m / s$
At a slope of displacement time graph gives the velocity of particle at that instant.
slope of the graph is zero at $t = 4 s$
then, the velocity is zero at $t = 4 s$
Hence, $V_{a v} = - 1 m / s, V = 0 m / s$

TS EAMCET(Medical)-2017

Motion in One Dimensions

141255 The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
original image

1 The total distance travelled by the particle is zero

2 The total displacement of the particle is zero

3 The maximum acceleration of the particle is

\frac{1}{2} {ms}^{- 2}

4 The total distance travelled by the particle at the end of

10 s

is

100 m

5 At the fifth second, the acceleration of the particle is

2 {ms}^{- 2}

Explanation:

B $v = \frac{ds}{dt}$
$\int ds = \int vdt$
$Δ s =$ Area under v-t curve
$Δ s = d_{1} + d_{2}$
$Δ s = \frac{1}{2} \times 10 \times 10 - \frac{1}{2} \times 10 \times 10$
$∴ Δ s = 0$

Kerala CEE -2018

Motion in One Dimensions

141257 A car starts from rest with a constant acceleration of $5 {ms}^{- 2}$ . At the same time a bus travelling with uniform velocity of $50 {ms}^{- 1}$ overtakes and passes the car. The distance through which the car again overtakes the bus is:

1

250 m

2

500 m

3

750 m

4

1000 m

Explanation:

D Given, $s = 50 t, u = 0, a = 5 m / s^{2}$
Distance covered by bus will be equal to distance covered by car.
$∵ s = ut + \frac{1}{2} {at}^{2}$
$50 t = 0 \times t + \frac{1}{2} 5 t^{2}$
$t = 20 \sec$
Distance $=$ speed $\times$ time $= 50 \times 20 = 1000 m$

Assam CEE-2017

Motion in One Dimensions

141258 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

1

2

3

4

Explanation:

B $v = u - gt = u - 10 t (∵ g = 10 m / s^{2})$ $v - t$ graph
original image
$y = u - 10 x$ (straight line)
$y = c + mx$
compare-
Slope $(m) = - 10 (- ve)$
$c = u (+ ve)$

JEE Main 2017

Motion in One Dimensions

141259 The displacement of a particle moving along $x$ axis versus time is given in the figure below: The average velocity $(V_{avg})$ of the particle in the first four seconds and the velocity $V$ of it at $t = 4 s$ are
original image

1

V_{arg} = - 1 m / s, V = - 1 m / s

2

V_{avg} = 1 m / s, V = 1 m / s

3

V_{avg} = 1 m / s, V = 0 m / s

4

V_{avg} = - 1 m / s, V = 0 m / s

Explanation:

D Let us consider
As $t = 0$ , particle is at a position $5 m$ , and at $t = 4 s$ particle is at a position $1 m$ .
Displacement of particle $(s) = 1 - 5 = - 4 m$ time taken $(t) = 4 s$
original image
Average velocity $(V_{avg}) = \frac{s}{t} = \frac{- 4}{4} = - 1 m / s$
At a slope of displacement time graph gives the velocity of particle at that instant.
slope of the graph is zero at $t = 4 s$
then, the velocity is zero at $t = 4 s$
Hence, $V_{a v} = - 1 m / s, V = 0 m / s$

TS EAMCET(Medical)-2017

Motion in One Dimensions

141255 The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
original image

1 The total distance travelled by the particle is zero

2 The total displacement of the particle is zero

3 The maximum acceleration of the particle is

\frac{1}{2} {ms}^{- 2}

4 The total distance travelled by the particle at the end of

10 s

is

100 m

5 At the fifth second, the acceleration of the particle is

2 {ms}^{- 2}

Explanation:

B $v = \frac{ds}{dt}$
$\int ds = \int vdt$
$Δ s =$ Area under v-t curve
$Δ s = d_{1} + d_{2}$
$Δ s = \frac{1}{2} \times 10 \times 10 - \frac{1}{2} \times 10 \times 10$
$∴ Δ s = 0$

Kerala CEE -2018

Motion in One Dimensions

141257 A car starts from rest with a constant acceleration of $5 {ms}^{- 2}$ . At the same time a bus travelling with uniform velocity of $50 {ms}^{- 1}$ overtakes and passes the car. The distance through which the car again overtakes the bus is:

1

250 m

2

500 m

3

750 m

4

1000 m

Explanation:

D Given, $s = 50 t, u = 0, a = 5 m / s^{2}$
Distance covered by bus will be equal to distance covered by car.
$∵ s = ut + \frac{1}{2} {at}^{2}$
$50 t = 0 \times t + \frac{1}{2} 5 t^{2}$
$t = 20 \sec$
Distance $=$ speed $\times$ time $= 50 \times 20 = 1000 m$

Assam CEE-2017

Motion in One Dimensions

141258 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

1

2

3

4

Explanation:

B $v = u - gt = u - 10 t (∵ g = 10 m / s^{2})$ $v - t$ graph
original image
$y = u - 10 x$ (straight line)
$y = c + mx$
compare-
Slope $(m) = - 10 (- ve)$
$c = u (+ ve)$

JEE Main 2017

Motion in One Dimensions

141259 The displacement of a particle moving along $x$ axis versus time is given in the figure below: The average velocity $(V_{avg})$ of the particle in the first four seconds and the velocity $V$ of it at $t = 4 s$ are
original image

1

V_{arg} = - 1 m / s, V = - 1 m / s

2

V_{avg} = 1 m / s, V = 1 m / s

3

V_{avg} = 1 m / s, V = 0 m / s

4

V_{avg} = - 1 m / s, V = 0 m / s

Explanation:

D Let us consider
As $t = 0$ , particle is at a position $5 m$ , and at $t = 4 s$ particle is at a position $1 m$ .
Displacement of particle $(s) = 1 - 5 = - 4 m$ time taken $(t) = 4 s$
original image
Average velocity $(V_{avg}) = \frac{s}{t} = \frac{- 4}{4} = - 1 m / s$
At a slope of displacement time graph gives the velocity of particle at that instant.
slope of the graph is zero at $t = 4 s$
then, the velocity is zero at $t = 4 s$
Hence, $V_{a v} = - 1 m / s, V = 0 m / s$

TS EAMCET(Medical)-2017

Motion in One Dimensions

141255 The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
original image

1 The total distance travelled by the particle is zero

2 The total displacement of the particle is zero

3 The maximum acceleration of the particle is

\frac{1}{2} {ms}^{- 2}

4 The total distance travelled by the particle at the end of

10 s

is

100 m

5 At the fifth second, the acceleration of the particle is

2 {ms}^{- 2}

Explanation:

B $v = \frac{ds}{dt}$
$\int ds = \int vdt$
$Δ s =$ Area under v-t curve
$Δ s = d_{1} + d_{2}$
$Δ s = \frac{1}{2} \times 10 \times 10 - \frac{1}{2} \times 10 \times 10$
$∴ Δ s = 0$

Kerala CEE -2018

Motion in One Dimensions

141257 A car starts from rest with a constant acceleration of $5 {ms}^{- 2}$ . At the same time a bus travelling with uniform velocity of $50 {ms}^{- 1}$ overtakes and passes the car. The distance through which the car again overtakes the bus is:

1

250 m

2

500 m

3

750 m

4

1000 m

Explanation:

D Given, $s = 50 t, u = 0, a = 5 m / s^{2}$
Distance covered by bus will be equal to distance covered by car.
$∵ s = ut + \frac{1}{2} {at}^{2}$
$50 t = 0 \times t + \frac{1}{2} 5 t^{2}$
$t = 20 \sec$
Distance $=$ speed $\times$ time $= 50 \times 20 = 1000 m$

Assam CEE-2017

Motion in One Dimensions

141258 A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

1

2

3

4

Explanation:

B $v = u - gt = u - 10 t (∵ g = 10 m / s^{2})$ $v - t$ graph
original image
$y = u - 10 x$ (straight line)
$y = c + mx$
compare-
Slope $(m) = - 10 (- ve)$
$c = u (+ ve)$

JEE Main 2017

Motion in One Dimensions

141259 The displacement of a particle moving along $x$ axis versus time is given in the figure below: The average velocity $(V_{avg})$ of the particle in the first four seconds and the velocity $V$ of it at $t = 4 s$ are
original image

1

V_{arg} = - 1 m / s, V = - 1 m / s

2

V_{avg} = 1 m / s, V = 1 m / s

3

V_{avg} = 1 m / s, V = 0 m / s

4

V_{avg} = - 1 m / s, V = 0 m / s

Explanation:

D Let us consider
As $t = 0$ , particle is at a position $5 m$ , and at $t = 4 s$ particle is at a position $1 m$ .
Displacement of particle $(s) = 1 - 5 = - 4 m$ time taken $(t) = 4 s$
original image
Average velocity $(V_{avg}) = \frac{s}{t} = \frac{- 4}{4} = - 1 m / s$
At a slope of displacement time graph gives the velocity of particle at that instant.
slope of the graph is zero at $t = 4 s$
then, the velocity is zero at $t = 4 s$
Hence, $V_{a v} = - 1 m / s, V = 0 m / s$

TS EAMCET(Medical)-2017