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PHXI03:MOTION IN A STRAIGHT LINE

362319 A metro train starts from rest and in $5 s$ achieves $180 k m h^{- 1}$ After that it moves with constant velocity and comes to rest after travelling $45 m$ with uniform retardation. If total distance travelled is $395 m,$ find total time of travelling.

1

12.2 s

2

15.3 s

3

9 s

4

17.2 s

Explanation:

Given, $v = 108 k m h^{- 1} = 30 m s^{- 1}$ and $t = 5 s$
From first equation of motion,
$v = u + a t$
$\Rightarrow 30 = 0 + a \times 5$
$(∴ u = 0) \Rightarrow a = 6 m s^{- 2}$
So, distance travelled by metro train in $5 s,$
$s_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times (6) \times (5)^{2} = 75 m$
Distance travelled before coming to rest $= 45 m$
So, from third equation of motion,
$0^{2} = (30)^{2} - 2 a^{'} \times 45 a^{'}$
$a^{'} = \frac{30 \times 30}{2 \times 45} = 10 m s^{- 2}$
Time taken in travelling $45 m,$
$t_{3} = \frac{30}{10} = 3 s$
Now, total distance $= 395 m$
i.e., $75 + s^{'} + 45 = 395 m$
$\Rightarrow s^{'} = 395 - (75 + 45) = 275 m$
$∴ t_{2} = \frac{275}{30} = 9.2 s$
Hence, total time taken in whole journey
$= t_{1} + t_{2} + t_{3}$
$= 5 + 9.2 + 3 = 17.2 s$

PHXI03:MOTION IN A STRAIGHT LINE

362320 A truck is moving at a speed of $72 {k m h}^{- 1}$ on a straight road. The driver can produce deceleration of $2 {m s}^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is 0.2 s . The value of $x$ is

1 5

2 10

3 8

4 2

Explanation:

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = {(72 \times \frac{5}{18})}^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance $= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$

PHXI03:MOTION IN A STRAIGHT LINE

362321 Assertion :
The displacement -time graph of a body moving with uniform acceleration is a straight line.
Reason :
The displacement is proportional to time for uniform motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The displacement of a body moving in straight line with constant acceleration is given by, $s = u t + \frac{1}{2} a t^{2}$ .
This is a equation of a parabola, not straight line. Therefore the displacement-time graph is a parabola. The displacement time graph will be straight line, if and only if acceleration of body is zero or body is moving with uniform velocity. Uniform motion means constant velocity. So correct option is (4).

PHXI03:MOTION IN A STRAIGHT LINE

362322 A car, starting from rest, accelerates at the rate $f$ through a distance $S$ , then continues at constant speed for time $t$ and then decelerates at the rates $\frac{f}{2}$ to come to rest. If the total distance travelled is 15 $S$ , then

1

S = \frac{1}{6} f t^{2}

2

S = \frac{1}{72} f t^{2}

3

S = \frac{1}{2} f t^{2}

4

S = \frac{1}{4} f t^{2}

Explanation:

Let car start from point $A$ from rest and moves up to point $B$ with acceleration $f$
Velocity of car at point $B$ , $υ = \sqrt{2 f S}$
$[A s υ^{2} = u^{2} + 2 a s]$
Car moves distance $B C$ with this constant velocity in time $t$
$x = \sqrt{2 f S} . t [A s s = u t]$ (1)
So the velocity of car at point $C$ also will be
$\sqrt{2 f S}$ and finally car stops after covering distance $y$ .
Distance $C D \Rightarrow y = \frac{{(\sqrt{2 f S})}^{2}}{2 (f / 2)} = \frac{2 f S}{f} = 2 S$ (2)
So, the total distance $A D = A B + B C + C D = 45 S$ [Given]
$\Rightarrow S + x + 2 S = 15 S \Rightarrow x = 12 S$
Substituting the values of $x$ in equation (1) we get
$x = \sqrt{2 f S} . t \Rightarrow 12 S = \sqrt{2 f S} . t \Rightarrow 144 S^{2} = 2 f S . t^{2}$
$\Rightarrow S = \frac{1}{72} f t^{2}$ .

PHXI03:MOTION IN A STRAIGHT LINE

362319 A metro train starts from rest and in $5 s$ achieves $180 k m h^{- 1}$ After that it moves with constant velocity and comes to rest after travelling $45 m$ with uniform retardation. If total distance travelled is $395 m,$ find total time of travelling.

1

12.2 s

2

15.3 s

3

9 s

4

17.2 s

Explanation:

Given, $v = 108 k m h^{- 1} = 30 m s^{- 1}$ and $t = 5 s$
From first equation of motion,
$v = u + a t$
$\Rightarrow 30 = 0 + a \times 5$
$(∴ u = 0) \Rightarrow a = 6 m s^{- 2}$
So, distance travelled by metro train in $5 s,$
$s_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times (6) \times (5)^{2} = 75 m$
Distance travelled before coming to rest $= 45 m$
So, from third equation of motion,
$0^{2} = (30)^{2} - 2 a^{'} \times 45 a^{'}$
$a^{'} = \frac{30 \times 30}{2 \times 45} = 10 m s^{- 2}$
Time taken in travelling $45 m,$
$t_{3} = \frac{30}{10} = 3 s$
Now, total distance $= 395 m$
i.e., $75 + s^{'} + 45 = 395 m$
$\Rightarrow s^{'} = 395 - (75 + 45) = 275 m$
$∴ t_{2} = \frac{275}{30} = 9.2 s$
Hence, total time taken in whole journey
$= t_{1} + t_{2} + t_{3}$
$= 5 + 9.2 + 3 = 17.2 s$

PHXI03:MOTION IN A STRAIGHT LINE

362320 A truck is moving at a speed of $72 {k m h}^{- 1}$ on a straight road. The driver can produce deceleration of $2 {m s}^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is 0.2 s . The value of $x$ is

1 5

2 10

3 8

4 2

Explanation:

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = {(72 \times \frac{5}{18})}^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance $= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$

PHXI03:MOTION IN A STRAIGHT LINE

362321 Assertion :
The displacement -time graph of a body moving with uniform acceleration is a straight line.
Reason :
The displacement is proportional to time for uniform motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The displacement of a body moving in straight line with constant acceleration is given by, $s = u t + \frac{1}{2} a t^{2}$ .
This is a equation of a parabola, not straight line. Therefore the displacement-time graph is a parabola. The displacement time graph will be straight line, if and only if acceleration of body is zero or body is moving with uniform velocity. Uniform motion means constant velocity. So correct option is (4).

PHXI03:MOTION IN A STRAIGHT LINE

362322 A car, starting from rest, accelerates at the rate $f$ through a distance $S$ , then continues at constant speed for time $t$ and then decelerates at the rates $\frac{f}{2}$ to come to rest. If the total distance travelled is 15 $S$ , then

1

S = \frac{1}{6} f t^{2}

2

S = \frac{1}{72} f t^{2}

3

S = \frac{1}{2} f t^{2}

4

S = \frac{1}{4} f t^{2}

Explanation:

Let car start from point $A$ from rest and moves up to point $B$ with acceleration $f$
Velocity of car at point $B$ , $υ = \sqrt{2 f S}$
$[A s υ^{2} = u^{2} + 2 a s]$
Car moves distance $B C$ with this constant velocity in time $t$
$x = \sqrt{2 f S} . t [A s s = u t]$ (1)
So the velocity of car at point $C$ also will be
$\sqrt{2 f S}$ and finally car stops after covering distance $y$ .
Distance $C D \Rightarrow y = \frac{{(\sqrt{2 f S})}^{2}}{2 (f / 2)} = \frac{2 f S}{f} = 2 S$ (2)
So, the total distance $A D = A B + B C + C D = 45 S$ [Given]
$\Rightarrow S + x + 2 S = 15 S \Rightarrow x = 12 S$
Substituting the values of $x$ in equation (1) we get
$x = \sqrt{2 f S} . t \Rightarrow 12 S = \sqrt{2 f S} . t \Rightarrow 144 S^{2} = 2 f S . t^{2}$
$\Rightarrow S = \frac{1}{72} f t^{2}$ .

PHXI03:MOTION IN A STRAIGHT LINE

362319 A metro train starts from rest and in $5 s$ achieves $180 k m h^{- 1}$ After that it moves with constant velocity and comes to rest after travelling $45 m$ with uniform retardation. If total distance travelled is $395 m,$ find total time of travelling.

1

12.2 s

2

15.3 s

3

9 s

4

17.2 s

Explanation:

Given, $v = 108 k m h^{- 1} = 30 m s^{- 1}$ and $t = 5 s$
From first equation of motion,
$v = u + a t$
$\Rightarrow 30 = 0 + a \times 5$
$(∴ u = 0) \Rightarrow a = 6 m s^{- 2}$
So, distance travelled by metro train in $5 s,$
$s_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times (6) \times (5)^{2} = 75 m$
Distance travelled before coming to rest $= 45 m$
So, from third equation of motion,
$0^{2} = (30)^{2} - 2 a^{'} \times 45 a^{'}$
$a^{'} = \frac{30 \times 30}{2 \times 45} = 10 m s^{- 2}$
Time taken in travelling $45 m,$
$t_{3} = \frac{30}{10} = 3 s$
Now, total distance $= 395 m$
i.e., $75 + s^{'} + 45 = 395 m$
$\Rightarrow s^{'} = 395 - (75 + 45) = 275 m$
$∴ t_{2} = \frac{275}{30} = 9.2 s$
Hence, total time taken in whole journey
$= t_{1} + t_{2} + t_{3}$
$= 5 + 9.2 + 3 = 17.2 s$

PHXI03:MOTION IN A STRAIGHT LINE

362320 A truck is moving at a speed of $72 {k m h}^{- 1}$ on a straight road. The driver can produce deceleration of $2 {m s}^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is 0.2 s . The value of $x$ is

1 5

2 10

3 8

4 2

Explanation:

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = {(72 \times \frac{5}{18})}^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance $= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$

PHXI03:MOTION IN A STRAIGHT LINE

362321 Assertion :
The displacement -time graph of a body moving with uniform acceleration is a straight line.
Reason :
The displacement is proportional to time for uniform motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The displacement of a body moving in straight line with constant acceleration is given by, $s = u t + \frac{1}{2} a t^{2}$ .
This is a equation of a parabola, not straight line. Therefore the displacement-time graph is a parabola. The displacement time graph will be straight line, if and only if acceleration of body is zero or body is moving with uniform velocity. Uniform motion means constant velocity. So correct option is (4).

PHXI03:MOTION IN A STRAIGHT LINE

362322 A car, starting from rest, accelerates at the rate $f$ through a distance $S$ , then continues at constant speed for time $t$ and then decelerates at the rates $\frac{f}{2}$ to come to rest. If the total distance travelled is 15 $S$ , then

1

S = \frac{1}{6} f t^{2}

2

S = \frac{1}{72} f t^{2}

3

S = \frac{1}{2} f t^{2}

4

S = \frac{1}{4} f t^{2}

Explanation:

Let car start from point $A$ from rest and moves up to point $B$ with acceleration $f$
Velocity of car at point $B$ , $υ = \sqrt{2 f S}$
$[A s υ^{2} = u^{2} + 2 a s]$
Car moves distance $B C$ with this constant velocity in time $t$
$x = \sqrt{2 f S} . t [A s s = u t]$ (1)
So the velocity of car at point $C$ also will be
$\sqrt{2 f S}$ and finally car stops after covering distance $y$ .
Distance $C D \Rightarrow y = \frac{{(\sqrt{2 f S})}^{2}}{2 (f / 2)} = \frac{2 f S}{f} = 2 S$ (2)
So, the total distance $A D = A B + B C + C D = 45 S$ [Given]
$\Rightarrow S + x + 2 S = 15 S \Rightarrow x = 12 S$
Substituting the values of $x$ in equation (1) we get
$x = \sqrt{2 f S} . t \Rightarrow 12 S = \sqrt{2 f S} . t \Rightarrow 144 S^{2} = 2 f S . t^{2}$
$\Rightarrow S = \frac{1}{72} f t^{2}$ .

PHXI03:MOTION IN A STRAIGHT LINE

362319 A metro train starts from rest and in $5 s$ achieves $180 k m h^{- 1}$ After that it moves with constant velocity and comes to rest after travelling $45 m$ with uniform retardation. If total distance travelled is $395 m,$ find total time of travelling.

1

12.2 s

2

15.3 s

3

9 s

4

17.2 s

Explanation:

Given, $v = 108 k m h^{- 1} = 30 m s^{- 1}$ and $t = 5 s$
From first equation of motion,
$v = u + a t$
$\Rightarrow 30 = 0 + a \times 5$
$(∴ u = 0) \Rightarrow a = 6 m s^{- 2}$
So, distance travelled by metro train in $5 s,$
$s_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times (6) \times (5)^{2} = 75 m$
Distance travelled before coming to rest $= 45 m$
So, from third equation of motion,
$0^{2} = (30)^{2} - 2 a^{'} \times 45 a^{'}$
$a^{'} = \frac{30 \times 30}{2 \times 45} = 10 m s^{- 2}$
Time taken in travelling $45 m,$
$t_{3} = \frac{30}{10} = 3 s$
Now, total distance $= 395 m$
i.e., $75 + s^{'} + 45 = 395 m$
$\Rightarrow s^{'} = 395 - (75 + 45) = 275 m$
$∴ t_{2} = \frac{275}{30} = 9.2 s$
Hence, total time taken in whole journey
$= t_{1} + t_{2} + t_{3}$
$= 5 + 9.2 + 3 = 17.2 s$

PHXI03:MOTION IN A STRAIGHT LINE

362320 A truck is moving at a speed of $72 {k m h}^{- 1}$ on a straight road. The driver can produce deceleration of $2 {m s}^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is 0.2 s . The value of $x$ is

1 5

2 10

3 8

4 2

Explanation:

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = {(72 \times \frac{5}{18})}^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance $= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$

PHXI03:MOTION IN A STRAIGHT LINE

362321 Assertion :
The displacement -time graph of a body moving with uniform acceleration is a straight line.
Reason :
The displacement is proportional to time for uniform motion.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

The displacement of a body moving in straight line with constant acceleration is given by, $s = u t + \frac{1}{2} a t^{2}$ .
This is a equation of a parabola, not straight line. Therefore the displacement-time graph is a parabola. The displacement time graph will be straight line, if and only if acceleration of body is zero or body is moving with uniform velocity. Uniform motion means constant velocity. So correct option is (4).

PHXI03:MOTION IN A STRAIGHT LINE

362322 A car, starting from rest, accelerates at the rate $f$ through a distance $S$ , then continues at constant speed for time $t$ and then decelerates at the rates $\frac{f}{2}$ to come to rest. If the total distance travelled is 15 $S$ , then

1

S = \frac{1}{6} f t^{2}

2

S = \frac{1}{72} f t^{2}

3

S = \frac{1}{2} f t^{2}

4

S = \frac{1}{4} f t^{2}

Explanation:

Let car start from point $A$ from rest and moves up to point $B$ with acceleration $f$
Velocity of car at point $B$ , $υ = \sqrt{2 f S}$
$[A s υ^{2} = u^{2} + 2 a s]$
Car moves distance $B C$ with this constant velocity in time $t$
$x = \sqrt{2 f S} . t [A s s = u t]$ (1)
So the velocity of car at point $C$ also will be
$\sqrt{2 f S}$ and finally car stops after covering distance $y$ .
Distance $C D \Rightarrow y = \frac{{(\sqrt{2 f S})}^{2}}{2 (f / 2)} = \frac{2 f S}{f} = 2 S$ (2)
So, the total distance $A D = A B + B C + C D = 45 S$ [Given]
$\Rightarrow S + x + 2 S = 15 S \Rightarrow x = 12 S$
Substituting the values of $x$ in equation (1) we get
$x = \sqrt{2 f S} . t \Rightarrow 12 S = \sqrt{2 f S} . t \Rightarrow 144 S^{2} = 2 f S . t^{2}$
$\Rightarrow S = \frac{1}{72} f t^{2}$ .

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