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Motion in One Dimensions

141292 In the following velocity-time graph, the distance travelled by the body, in metre is
original image

1 200

2 250

3 300

4 400

Explanation:

B From given figure,
original image
Distance travelled by the body is $S =$ Area of $△ OAE +$ Area of $◻ EDAB +$ Area of $△ DBC$
$= \frac{1}{2} \times 10 \times 10 + 10 \times 10 + \frac{1}{2} \times 10 \times 20$
$= 50 + 100 + 100$
$= 250 m$

EAMCET-1994

Motion in One Dimensions

141293 A body A starts from rest with an acceleration $a_{1}$ . After $2 s$ another body $B$ starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5th second after the start of $A$ .
$a_{1} : a_{2}$ is equal to

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

A As per given condition,
Distance travelled in $5^{th}$ second $=$ distance travelled in $3^{rd}$ second.
$u + \frac{a_{1}}{2} (2 n - 1) = u + \frac{a_{2}}{2} (2 n - 1)$
Or, $\frac{a_{1}}{2} (2 \times 5 - 1) = \frac{a_{2}}{2} (2 \times 3 - 1) [∵ u = 0]$
Or, $\frac{9}{2} a_{1} = \frac{5}{2} \cdot a_{2}$
Or, $\frac{a_{1}}{a_{2}} = \frac{5}{9}$
$∴ a_{1} : a_{2} = 5 : 9$

EAMCET-1994

Motion in One Dimensions

141294 A body falling for $2 s$ covers a distance $s$ which is equal to that covered in next $1 s$ . If $g = 10 {ms}^{- 2}$ , the distance $s$ is

1

30 m

2

10 m

3

60 m

4

20 m

Explanation:

A As per given condition,
If $u =$ initial velocity,
Distance covered in $2 \sec =$ distance covered in $3^{rd} \sec$
Distance,
$S = ut + \frac{1}{2} {gt}^{2} = u + \frac{g}{2} (2 n - 1)$
$u \times 2 + \frac{1}{2} \times 10 \times 2^{2} = u + \frac{10}{2} (2 \times 3 - 1)$
$2 u + 20 = u + 25$
$u = 5 m / s$
$s = ut + \frac{1}{2} {gt}^{2}$
$s = 5 \times 2 + \frac{1}{2} \times 10 \times 2^{2}$
$s = 10 + 20$
$s = 30 m$

EAMCET-1995

Motion in One Dimensions

141296 The reaction time for a car diver is $0.9 s$ . If the car travelling initially with $36 km h^{- 1}$ is stopped by the driver after observing a signal by the deceleration of $5 {ms}^{- 2}$ , the total distance travelled by the car before coming to rest is

1

19 m

2

9 m

3

10 m

4

28 m

Explanation:

A Given data,
$u = 36 km / h = 36 \times \frac{5}{18} = 10 m / s$
$t = 0.9 \sec$
$s_{1} = ut = 10 \times 0.9 = 9 m \Rightarrow s_{1} = 9 m$
For deceleration, $v = u + gt$
$0 = 10 - 5 t_{1}$
$t_{1} = 2 \sec$
$∴$ Distance travelled after deceleration,
$s_{2} = {ut}_{1} - \frac{1}{2} {at}_{1}^{2}$
$Or, s_{2} = 10 \times 2 - \frac{1}{2} \times 5 \times 2^{2}$
$Or, s_{2} = 20 - 10$
$∴ s_{2} = 10 m$
So, total distance travelled by the car, $s = s_{1} + s_{2}$
$s = 9 + 10 = 19$
$s = 19 m$

EAMCET-1997

Motion in One Dimensions

141292 In the following velocity-time graph, the distance travelled by the body, in metre is
original image

1 200

2 250

3 300

4 400

Explanation:

B From given figure,
original image
Distance travelled by the body is $S =$ Area of $△ OAE +$ Area of $◻ EDAB +$ Area of $△ DBC$
$= \frac{1}{2} \times 10 \times 10 + 10 \times 10 + \frac{1}{2} \times 10 \times 20$
$= 50 + 100 + 100$
$= 250 m$

EAMCET-1994

Motion in One Dimensions

141293 A body A starts from rest with an acceleration $a_{1}$ . After $2 s$ another body $B$ starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5th second after the start of $A$ .
$a_{1} : a_{2}$ is equal to

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

A As per given condition,
Distance travelled in $5^{th}$ second $=$ distance travelled in $3^{rd}$ second.
$u + \frac{a_{1}}{2} (2 n - 1) = u + \frac{a_{2}}{2} (2 n - 1)$
Or, $\frac{a_{1}}{2} (2 \times 5 - 1) = \frac{a_{2}}{2} (2 \times 3 - 1) [∵ u = 0]$
Or, $\frac{9}{2} a_{1} = \frac{5}{2} \cdot a_{2}$
Or, $\frac{a_{1}}{a_{2}} = \frac{5}{9}$
$∴ a_{1} : a_{2} = 5 : 9$

EAMCET-1994

Motion in One Dimensions

141294 A body falling for $2 s$ covers a distance $s$ which is equal to that covered in next $1 s$ . If $g = 10 {ms}^{- 2}$ , the distance $s$ is

1

30 m

2

10 m

3

60 m

4

20 m

Explanation:

A As per given condition,
If $u =$ initial velocity,
Distance covered in $2 \sec =$ distance covered in $3^{rd} \sec$
Distance,
$S = ut + \frac{1}{2} {gt}^{2} = u + \frac{g}{2} (2 n - 1)$
$u \times 2 + \frac{1}{2} \times 10 \times 2^{2} = u + \frac{10}{2} (2 \times 3 - 1)$
$2 u + 20 = u + 25$
$u = 5 m / s$
$s = ut + \frac{1}{2} {gt}^{2}$
$s = 5 \times 2 + \frac{1}{2} \times 10 \times 2^{2}$
$s = 10 + 20$
$s = 30 m$

EAMCET-1995

Motion in One Dimensions

141296 The reaction time for a car diver is $0.9 s$ . If the car travelling initially with $36 km h^{- 1}$ is stopped by the driver after observing a signal by the deceleration of $5 {ms}^{- 2}$ , the total distance travelled by the car before coming to rest is

1

19 m

2

9 m

3

10 m

4

28 m

Explanation:

A Given data,
$u = 36 km / h = 36 \times \frac{5}{18} = 10 m / s$
$t = 0.9 \sec$
$s_{1} = ut = 10 \times 0.9 = 9 m \Rightarrow s_{1} = 9 m$
For deceleration, $v = u + gt$
$0 = 10 - 5 t_{1}$
$t_{1} = 2 \sec$
$∴$ Distance travelled after deceleration,
$s_{2} = {ut}_{1} - \frac{1}{2} {at}_{1}^{2}$
$Or, s_{2} = 10 \times 2 - \frac{1}{2} \times 5 \times 2^{2}$
$Or, s_{2} = 20 - 10$
$∴ s_{2} = 10 m$
So, total distance travelled by the car, $s = s_{1} + s_{2}$
$s = 9 + 10 = 19$
$s = 19 m$

EAMCET-1997

Motion in One Dimensions

141292 In the following velocity-time graph, the distance travelled by the body, in metre is
original image

1 200

2 250

3 300

4 400

Explanation:

B From given figure,
original image
Distance travelled by the body is $S =$ Area of $△ OAE +$ Area of $◻ EDAB +$ Area of $△ DBC$
$= \frac{1}{2} \times 10 \times 10 + 10 \times 10 + \frac{1}{2} \times 10 \times 20$
$= 50 + 100 + 100$
$= 250 m$

EAMCET-1994

Motion in One Dimensions

141293 A body A starts from rest with an acceleration $a_{1}$ . After $2 s$ another body $B$ starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5th second after the start of $A$ .
$a_{1} : a_{2}$ is equal to

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

A As per given condition,
Distance travelled in $5^{th}$ second $=$ distance travelled in $3^{rd}$ second.
$u + \frac{a_{1}}{2} (2 n - 1) = u + \frac{a_{2}}{2} (2 n - 1)$
Or, $\frac{a_{1}}{2} (2 \times 5 - 1) = \frac{a_{2}}{2} (2 \times 3 - 1) [∵ u = 0]$
Or, $\frac{9}{2} a_{1} = \frac{5}{2} \cdot a_{2}$
Or, $\frac{a_{1}}{a_{2}} = \frac{5}{9}$
$∴ a_{1} : a_{2} = 5 : 9$

EAMCET-1994

Motion in One Dimensions

141294 A body falling for $2 s$ covers a distance $s$ which is equal to that covered in next $1 s$ . If $g = 10 {ms}^{- 2}$ , the distance $s$ is

1

30 m

2

10 m

3

60 m

4

20 m

Explanation:

A As per given condition,
If $u =$ initial velocity,
Distance covered in $2 \sec =$ distance covered in $3^{rd} \sec$
Distance,
$S = ut + \frac{1}{2} {gt}^{2} = u + \frac{g}{2} (2 n - 1)$
$u \times 2 + \frac{1}{2} \times 10 \times 2^{2} = u + \frac{10}{2} (2 \times 3 - 1)$
$2 u + 20 = u + 25$
$u = 5 m / s$
$s = ut + \frac{1}{2} {gt}^{2}$
$s = 5 \times 2 + \frac{1}{2} \times 10 \times 2^{2}$
$s = 10 + 20$
$s = 30 m$

EAMCET-1995

Motion in One Dimensions

141296 The reaction time for a car diver is $0.9 s$ . If the car travelling initially with $36 km h^{- 1}$ is stopped by the driver after observing a signal by the deceleration of $5 {ms}^{- 2}$ , the total distance travelled by the car before coming to rest is

1

19 m

2

9 m

3

10 m

4

28 m

Explanation:

A Given data,
$u = 36 km / h = 36 \times \frac{5}{18} = 10 m / s$
$t = 0.9 \sec$
$s_{1} = ut = 10 \times 0.9 = 9 m \Rightarrow s_{1} = 9 m$
For deceleration, $v = u + gt$
$0 = 10 - 5 t_{1}$
$t_{1} = 2 \sec$
$∴$ Distance travelled after deceleration,
$s_{2} = {ut}_{1} - \frac{1}{2} {at}_{1}^{2}$
$Or, s_{2} = 10 \times 2 - \frac{1}{2} \times 5 \times 2^{2}$
$Or, s_{2} = 20 - 10$
$∴ s_{2} = 10 m$
So, total distance travelled by the car, $s = s_{1} + s_{2}$
$s = 9 + 10 = 19$
$s = 19 m$

EAMCET-1997

Motion in One Dimensions

141292 In the following velocity-time graph, the distance travelled by the body, in metre is
original image

1 200

2 250

3 300

4 400

Explanation:

B From given figure,
original image
Distance travelled by the body is $S =$ Area of $△ OAE +$ Area of $◻ EDAB +$ Area of $△ DBC$
$= \frac{1}{2} \times 10 \times 10 + 10 \times 10 + \frac{1}{2} \times 10 \times 20$
$= 50 + 100 + 100$
$= 250 m$

EAMCET-1994

Motion in One Dimensions

141293 A body A starts from rest with an acceleration $a_{1}$ . After $2 s$ another body $B$ starts from rest with an acceleration $a_{2}$ . If they travel equal distance in the 5th second after the start of $A$ .
$a_{1} : a_{2}$ is equal to

1

5 : 9

2

5 : 7

3

9 : 5

4

9 : 7

Explanation:

A As per given condition,
Distance travelled in $5^{th}$ second $=$ distance travelled in $3^{rd}$ second.
$u + \frac{a_{1}}{2} (2 n - 1) = u + \frac{a_{2}}{2} (2 n - 1)$
Or, $\frac{a_{1}}{2} (2 \times 5 - 1) = \frac{a_{2}}{2} (2 \times 3 - 1) [∵ u = 0]$
Or, $\frac{9}{2} a_{1} = \frac{5}{2} \cdot a_{2}$
Or, $\frac{a_{1}}{a_{2}} = \frac{5}{9}$
$∴ a_{1} : a_{2} = 5 : 9$

EAMCET-1994

Motion in One Dimensions

141294 A body falling for $2 s$ covers a distance $s$ which is equal to that covered in next $1 s$ . If $g = 10 {ms}^{- 2}$ , the distance $s$ is

1

30 m

2

10 m

3

60 m

4

20 m

Explanation:

A As per given condition,
If $u =$ initial velocity,
Distance covered in $2 \sec =$ distance covered in $3^{rd} \sec$
Distance,
$S = ut + \frac{1}{2} {gt}^{2} = u + \frac{g}{2} (2 n - 1)$
$u \times 2 + \frac{1}{2} \times 10 \times 2^{2} = u + \frac{10}{2} (2 \times 3 - 1)$
$2 u + 20 = u + 25$
$u = 5 m / s$
$s = ut + \frac{1}{2} {gt}^{2}$
$s = 5 \times 2 + \frac{1}{2} \times 10 \times 2^{2}$
$s = 10 + 20$
$s = 30 m$

EAMCET-1995

Motion in One Dimensions

141296 The reaction time for a car diver is $0.9 s$ . If the car travelling initially with $36 km h^{- 1}$ is stopped by the driver after observing a signal by the deceleration of $5 {ms}^{- 2}$ , the total distance travelled by the car before coming to rest is

1

19 m

2

9 m

3

10 m

4

28 m

Explanation:

A Given data,
$u = 36 km / h = 36 \times \frac{5}{18} = 10 m / s$
$t = 0.9 \sec$
$s_{1} = ut = 10 \times 0.9 = 9 m \Rightarrow s_{1} = 9 m$
For deceleration, $v = u + gt$
$0 = 10 - 5 t_{1}$
$t_{1} = 2 \sec$
$∴$ Distance travelled after deceleration,
$s_{2} = {ut}_{1} - \frac{1}{2} {at}_{1}^{2}$
$Or, s_{2} = 10 \times 2 - \frac{1}{2} \times 5 \times 2^{2}$
$Or, s_{2} = 20 - 10$
$∴ s_{2} = 10 m$
So, total distance travelled by the car, $s = s_{1} + s_{2}$
$s = 9 + 10 = 19$
$s = 19 m$

EAMCET-1997

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