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Motion in One Dimensions

141337 If the first one-third of a journey is travelled at $20 km / h$ , next one-third at $40 km / h$ and the last one-third at $60 km / h$ . The average speed of whole journey will be

1

32.7 km / h

2

35 km / h

3

40 km / h

4

45 km / h

Explanation:

B original image
$v = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{Total distance}{Total time}$
$= \frac{d}{\frac{d}{3 \times 20} + \frac{d}{3 \times 40} + \frac{d}{3 \times 60}}$
$= \frac{d}{\frac{d}{3 \times 20} [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}]}$
$= \frac{60}{1 + \frac{1}{2} + \frac{1}{3}} = \frac{60}{\frac{6 + 3 + 2}{6}}$
$= \frac{60 \times 6}{11} = \frac{360}{11}$
$v = 32.7 km / h$

UP CPMT-2002

Motion in One Dimensions

141339 A car accelerates from rest with $2 m / s^{2}$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100 m$ in 20 seconds. Then the maximum velocity attained by the car is

1

10 m / s

2

20 m / s

3

15 m / s

4

5 m / s

Explanation:

A By equation of motion,
$∵ v = u -$ at
$0 = u - a \times 20$
$u = 20 a$
and $v^{2} - u^{2} = 2 (- a) s$
$0^{2} - (20 a)^{2} = - 2 a \times 100$
$a = 0.5 m / s^{2}$
Now by equation (i),
$u = 20 a = 20 \times 0.5 = 10 m / s$

TS EAMCET (Engg.)-2016

Motion in One Dimensions

141340 A car starts from rest and starts moving on a circular path of radius $100 m$ such that its speed increases at the rate of $5 {ms}^{- 2}$ . What is the radial acceleration of the car at the instant it makes one complete round trip (in $m s^{- 2}$ )?

1

20 π

2

10 π

3

40 π

4

30 π

Explanation:

A The speed of the car after one complete round trip will be
$A = 5 {ms}^{- 2}$
$v^{2} = u^{2} + 2 as$
$v^{2} = 2 \times 5 (2 π r) [u = 0 and s = 2 π r)$
$v^{2} = 20 π r$
Radial acceleration of the particle will be
$a_{r} = \frac{v^{2}}{r} = \frac{20 π r}{r} = 20 π m / s^{2}$

SRM JEE - 2007

Motion in One Dimensions

141341 The position of a particle varies with time $t$ as $x$ $= a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time $t$ equal to

1

\frac{a}{b}

2

\frac{2 a}{b}

3

\frac{a}{3 b}

4 zero

Explanation:

C Position of a particle varies with time $t$ as $x = a t^{2} - b t^{3}$
Differentiating eq. (i) w.r.t. to time $t$ , we get velocity
$\frac{dx}{dt} = v = 2 at - 3 {bt}^{2}$
Differentiating eq. (ii) w.r.t. to time $t$ , we get acceleration
$\frac{dv}{dt} = 2 a - 6 bt$
For acceleration to be zero, $2 a - 6 b t = 0$
$\Rightarrow t = \frac{2 a}{6 b} = \frac{a}{3 b}$

SRM JEE - 2010

Motion in One Dimensions

141337 If the first one-third of a journey is travelled at $20 km / h$ , next one-third at $40 km / h$ and the last one-third at $60 km / h$ . The average speed of whole journey will be

1

32.7 km / h

2

35 km / h

3

40 km / h

4

45 km / h

Explanation:

B original image
$v = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{Total distance}{Total time}$
$= \frac{d}{\frac{d}{3 \times 20} + \frac{d}{3 \times 40} + \frac{d}{3 \times 60}}$
$= \frac{d}{\frac{d}{3 \times 20} [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}]}$
$= \frac{60}{1 + \frac{1}{2} + \frac{1}{3}} = \frac{60}{\frac{6 + 3 + 2}{6}}$
$= \frac{60 \times 6}{11} = \frac{360}{11}$
$v = 32.7 km / h$

UP CPMT-2002

Motion in One Dimensions

141339 A car accelerates from rest with $2 m / s^{2}$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100 m$ in 20 seconds. Then the maximum velocity attained by the car is

1

10 m / s

2

20 m / s

3

15 m / s

4

5 m / s

Explanation:

A By equation of motion,
$∵ v = u -$ at
$0 = u - a \times 20$
$u = 20 a$
and $v^{2} - u^{2} = 2 (- a) s$
$0^{2} - (20 a)^{2} = - 2 a \times 100$
$a = 0.5 m / s^{2}$
Now by equation (i),
$u = 20 a = 20 \times 0.5 = 10 m / s$

TS EAMCET (Engg.)-2016

Motion in One Dimensions

141340 A car starts from rest and starts moving on a circular path of radius $100 m$ such that its speed increases at the rate of $5 {ms}^{- 2}$ . What is the radial acceleration of the car at the instant it makes one complete round trip (in $m s^{- 2}$ )?

1

20 π

2

10 π

3

40 π

4

30 π

Explanation:

A The speed of the car after one complete round trip will be
$A = 5 {ms}^{- 2}$
$v^{2} = u^{2} + 2 as$
$v^{2} = 2 \times 5 (2 π r) [u = 0 and s = 2 π r)$
$v^{2} = 20 π r$
Radial acceleration of the particle will be
$a_{r} = \frac{v^{2}}{r} = \frac{20 π r}{r} = 20 π m / s^{2}$

SRM JEE - 2007

Motion in One Dimensions

141341 The position of a particle varies with time $t$ as $x$ $= a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time $t$ equal to

1

\frac{a}{b}

2

\frac{2 a}{b}

3

\frac{a}{3 b}

4 zero

Explanation:

C Position of a particle varies with time $t$ as $x = a t^{2} - b t^{3}$
Differentiating eq. (i) w.r.t. to time $t$ , we get velocity
$\frac{dx}{dt} = v = 2 at - 3 {bt}^{2}$
Differentiating eq. (ii) w.r.t. to time $t$ , we get acceleration
$\frac{dv}{dt} = 2 a - 6 bt$
For acceleration to be zero, $2 a - 6 b t = 0$
$\Rightarrow t = \frac{2 a}{6 b} = \frac{a}{3 b}$

SRM JEE - 2010

Motion in One Dimensions

141337 If the first one-third of a journey is travelled at $20 km / h$ , next one-third at $40 km / h$ and the last one-third at $60 km / h$ . The average speed of whole journey will be

1

32.7 km / h

2

35 km / h

3

40 km / h

4

45 km / h

Explanation:

B original image
$v = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{Total distance}{Total time}$
$= \frac{d}{\frac{d}{3 \times 20} + \frac{d}{3 \times 40} + \frac{d}{3 \times 60}}$
$= \frac{d}{\frac{d}{3 \times 20} [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}]}$
$= \frac{60}{1 + \frac{1}{2} + \frac{1}{3}} = \frac{60}{\frac{6 + 3 + 2}{6}}$
$= \frac{60 \times 6}{11} = \frac{360}{11}$
$v = 32.7 km / h$

UP CPMT-2002

Motion in One Dimensions

141339 A car accelerates from rest with $2 m / s^{2}$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100 m$ in 20 seconds. Then the maximum velocity attained by the car is

1

10 m / s

2

20 m / s

3

15 m / s

4

5 m / s

Explanation:

A By equation of motion,
$∵ v = u -$ at
$0 = u - a \times 20$
$u = 20 a$
and $v^{2} - u^{2} = 2 (- a) s$
$0^{2} - (20 a)^{2} = - 2 a \times 100$
$a = 0.5 m / s^{2}$
Now by equation (i),
$u = 20 a = 20 \times 0.5 = 10 m / s$

TS EAMCET (Engg.)-2016

Motion in One Dimensions

141340 A car starts from rest and starts moving on a circular path of radius $100 m$ such that its speed increases at the rate of $5 {ms}^{- 2}$ . What is the radial acceleration of the car at the instant it makes one complete round trip (in $m s^{- 2}$ )?

1

20 π

2

10 π

3

40 π

4

30 π

Explanation:

A The speed of the car after one complete round trip will be
$A = 5 {ms}^{- 2}$
$v^{2} = u^{2} + 2 as$
$v^{2} = 2 \times 5 (2 π r) [u = 0 and s = 2 π r)$
$v^{2} = 20 π r$
Radial acceleration of the particle will be
$a_{r} = \frac{v^{2}}{r} = \frac{20 π r}{r} = 20 π m / s^{2}$

SRM JEE - 2007

Motion in One Dimensions

141341 The position of a particle varies with time $t$ as $x$ $= a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time $t$ equal to

1

\frac{a}{b}

2

\frac{2 a}{b}

3

\frac{a}{3 b}

4 zero

Explanation:

C Position of a particle varies with time $t$ as $x = a t^{2} - b t^{3}$
Differentiating eq. (i) w.r.t. to time $t$ , we get velocity
$\frac{dx}{dt} = v = 2 at - 3 {bt}^{2}$
Differentiating eq. (ii) w.r.t. to time $t$ , we get acceleration
$\frac{dv}{dt} = 2 a - 6 bt$
For acceleration to be zero, $2 a - 6 b t = 0$
$\Rightarrow t = \frac{2 a}{6 b} = \frac{a}{3 b}$

SRM JEE - 2010

Motion in One Dimensions

141337 If the first one-third of a journey is travelled at $20 km / h$ , next one-third at $40 km / h$ and the last one-third at $60 km / h$ . The average speed of whole journey will be

1

32.7 km / h

2

35 km / h

3

40 km / h

4

45 km / h

Explanation:

B original image
$v = \frac{d}{t_{1} + t_{2} + t_{3}} = \frac{Total distance}{Total time}$
$= \frac{d}{\frac{d}{3 \times 20} + \frac{d}{3 \times 40} + \frac{d}{3 \times 60}}$
$= \frac{d}{\frac{d}{3 \times 20} [\frac{1}{1} + \frac{1}{2} + \frac{1}{3}]}$
$= \frac{60}{1 + \frac{1}{2} + \frac{1}{3}} = \frac{60}{\frac{6 + 3 + 2}{6}}$
$= \frac{60 \times 6}{11} = \frac{360}{11}$
$v = 32.7 km / h$

UP CPMT-2002

Motion in One Dimensions

141339 A car accelerates from rest with $2 m / s^{2}$ on a straight line path and then comes to rest after applying brakes. Total distance travelled by the car is $100 m$ in 20 seconds. Then the maximum velocity attained by the car is

1

10 m / s

2

20 m / s

3

15 m / s

4

5 m / s

Explanation:

A By equation of motion,
$∵ v = u -$ at
$0 = u - a \times 20$
$u = 20 a$
and $v^{2} - u^{2} = 2 (- a) s$
$0^{2} - (20 a)^{2} = - 2 a \times 100$
$a = 0.5 m / s^{2}$
Now by equation (i),
$u = 20 a = 20 \times 0.5 = 10 m / s$

TS EAMCET (Engg.)-2016

Motion in One Dimensions

141340 A car starts from rest and starts moving on a circular path of radius $100 m$ such that its speed increases at the rate of $5 {ms}^{- 2}$ . What is the radial acceleration of the car at the instant it makes one complete round trip (in $m s^{- 2}$ )?

1

20 π

2

10 π

3

40 π

4

30 π

Explanation:

A The speed of the car after one complete round trip will be
$A = 5 {ms}^{- 2}$
$v^{2} = u^{2} + 2 as$
$v^{2} = 2 \times 5 (2 π r) [u = 0 and s = 2 π r)$
$v^{2} = 20 π r$
Radial acceleration of the particle will be
$a_{r} = \frac{v^{2}}{r} = \frac{20 π r}{r} = 20 π m / s^{2}$

SRM JEE - 2007

Motion in One Dimensions

141341 The position of a particle varies with time $t$ as $x$ $= a t^{2} - b t^{3}$ . The acceleration of the particle will be zero at time $t$ equal to

1

\frac{a}{b}

2

\frac{2 a}{b}

3

\frac{a}{3 b}

4 zero

Explanation:

C Position of a particle varies with time $t$ as $x = a t^{2} - b t^{3}$
Differentiating eq. (i) w.r.t. to time $t$ , we get velocity
$\frac{dx}{dt} = v = 2 at - 3 {bt}^{2}$
Differentiating eq. (ii) w.r.t. to time $t$ , we get acceleration
$\frac{dv}{dt} = 2 a - 6 bt$
For acceleration to be zero, $2 a - 6 b t = 0$
$\Rightarrow t = \frac{2 a}{6 b} = \frac{a}{3 b}$

SRM JEE - 2010

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