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**Motion in Straight Line**

**Motion in Straight Line****Chapter: 2**

Part – I |

**EXERCISE**

**1. In which of the following examples of motion, can the body be considered approximately a point object:**

**(a) A railway carriage moving without jerks between two stations.**

Ans: The railway carriage moves without jerks between two stations, so, as compared to the size of the train the distance of the station is large. So, it can be considered a point object.

**(b) A monkey sitting on top of a man cycling smoothly on a circular track.**

Ans: The monkey and the man together are small compared to the entire circular track. As the distance covered by monkey is large in a reasonable time. So, they can be considered a point object.

**(c) A spinning cricket ball that turns sharply on hitting the ground.**

Ans: As turning of the ball is not smooth, Its rotation and subsequent motion involve its entire physical dimensions. The distance covered by the ball is not large in the reasonable time. Therefore the ball cannot be considered as a point object.

**(d) A tumbling beaker that has slipped off the edge of a table.**

Ans: A tumbling beaker that has slipped off the edge of a table, The size of the beaker is relevant to its tumbling motion. The distance covered by it is not large in reasonable time. Therefore the ball cannot be considered as a point object.

**2. The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below;**

**(a) (A/B) lives closer to the school than (B/A).**

Ans: A lives closer to school than B, because B has to cover higher distances as OP<OQ.

**(b) (A/B) starts from the school earlier than (B/A).**

Ans: A starts earlier from school B, because t = 0 for A but for B, t has some finite time.

**(c) (A/B) walks faster than (B/A).**

Ans: As slope of B is greater than that of A, thus B walks faster than A.

**(d) A and B reach home at the (same/different) time.**

Ans: A and B reach home at different times, as their graphs intersect the time axis at different points.

**(e) (A/B) overtakes (B/A) on the road (once/twice).**

Ans: A overtakes B on the road once, as seen from the intersection of their graphs where A’s position is greater than B’s position temporarily.

**3. A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.**

Ans: **Returning Home by Auto:**

**Speed:** 25 km/h

**Distance:** 2.5 km

**The distance covered while walking:** 2.5 km.

= 2.5/5

= 1/2h.

**Graph Segments:**

**Segment 1:** Walking to the Office (9:00 am – 9:30 am): Starts at (0, 0) and ends at (0.5, 2.5)

**Segment 2:** Staying at the Office (9:30 am – 5:00 pm): Starts at (0.5, 2.5) and ends at (8, 2.5)

**Segment 3:** Returning Home by Auto (5:00 pm – 5:06 pm): Starts at (8, 2.5) and ends at (8.1, 0)

**Now, time taken to return home:**

= 2.5/25

= 1/10 h

= 6 minutes.

So, at 5.06 p.m. x = 0. The graph shows a rapid decrease from 2.5 km to 0 km over 0.1 hours.

**4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.**

Ans: The graph shows that he takes 37 seconds to fall in the pit.

The distance of the pit from the start = 13m.

The time taken to move first is 5 m = 5 sec.

Each cycle consists of 5 steps forward (5 metres) followed by 3 steps backward (3 metres), resulting in a net movement of 2 metres forward every 8 seconds.

Time taken in moving 8m = 8 × 8 / 2

= 32 seconds.

The distance of the pit from the drunkard after covering a net distance of 8m = 13 – 8 = 5 sec.

**Now in the next 5 steps forward, he will take 5 sec more i.e. total time taken to fall in the pit:**

= 32 + 5 = 37 sec.

**5. A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?**

Ans: **Convert the speed from km/h to m/s:**

126 km/h = 126 × 1000/ 3600 m/s

= 35 m/s

Speed u = 35 m/s.

Initial speed = 35 m/s

**Final speed:** 0

Distance travelled (s) = 200 m.

As, the v^{2} – u^{2} = 2aS or 0 – (35)^{2} = 2a × 200

a = -(35)^{2} / 400

= 3.06 ms^{-2}

v = u + at

Since v = 0

0 = 35+(−3.0625)⋅t

3. 0625 t = 35

= t = 35/ 3. 0625

= 11. 4 sec.

**6. A player throws a ball upwards with an initial speed of 29.4 m s**^{–1}**.**

**(a) What is the direction of acceleration during the upward motion of the ball?**

Ans: When the ball is thrown upwards, gravity acts downwards (towards the Earth). The ball is under acceleration due to gravity which always acts vertically downward. This is because gravity pulls everything towards the Earth’s centre.

**(b) What are the velocity and acceleration of the ball at the highest point of its motion?**

Ans: **Velocity and Acceleration at Highest Point:**

Velocity = Zero.

Acceleration = g = 9.8 m/s^{2}.

**(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.**

Ans:

For upward motion | |

(i) Position | Negative |

(ii) Velocity | Negative |

(iii) Acceleration | Positive |

For downward motion | |

(i) Position | Positive |

(ii) Velocity | Positive |

(iii) Acceleration | Positive |

**(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).**

Ans: H = V^{2}_{0} / 2g

Where V_{0} is the initial velocity (29.4 m/s) and g is the acceleration due to gravity (9.8 m/s^{2}).

H = (29. 4)2 / 2 × 9.8

H = 864.36 / 19.6

H = 44.1m/s.

So the ball height reaches approximately 44.1m/s.

**The total time ***T*** for the ball to go up and come back down can be found using:**

T = 2V_{0} / g

T = 2 × 29.4/ 9.8

T = 58.8 / 9.8

T = 6 sec.

**7. Read each statement below carefully and state with reasons and examples, if it is true or false; A particle in one-dimensional motion.**

**(a) With zero speed at an instant may have non-zero acceleration at that instant.**

Ans: True. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at each point during its journey.

**(b) With zero speed may have non-zero velocity**.

Ans: False. Because having speed means having magnitude of velocity.

**(c) With constant speed must have zero acceleration.**

Ans: True. Such a scenario is uniform circular motion of a particle about a centre. In such a case, acceleration is not zero.

**(d) With a positive value of acceleration must be speeding up.**

Ans: False. It can be true only when the chosen positive direction is along the direction of motion.

**8. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.**

Ans: The ball is dropped from a height h = 90 metres.

The acceleration due to gravity g = 9.8 m/s².

The initial velocity u = 0 m/s.

The equation of motion h = ½ gt2

= h = ½ gt^{2}

= 90 = ½ x 9.8 x t^{2}

= t^{2} = 180/9.8

t = 4. 29 s

**The velocity just before hitting the floor:**

v = gt

v = 9.8 × 4.29

= 42 m/s

**The ball loses one-tenth of its speed after the collision, so the speed after the first impact is:**

v^{1} = 0.9 × 42

= 37.8 m/s.

**9. Explain clearly, with examples, the distinction between: **

**(a) Magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval.**

Ans: The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.

The total path length of a particle is the actual path length covered by the particle in a given interval of time.

For example, the professor could pace back and forth many times, perhaps walking a distance of 150 metres during a lecture, yet still end up only two metres to the right of her starting point.

**(b) Magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only].**

Ans: The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. Velocity is the rate of change of displacement with respect to time. It is typically measured in metres per second (m/s) or miles per hour (mph).

The magnitude of average velocity is always less than or equal to the average speed because displacement is always smaller than or equal to distance. The average velocity of the particle can be positive as well as negative and its positive and negative value depends on the sign of displacement.

**10. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h**^{–1}**. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h**^{–1}**. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]**

Ans: To market: 2.5 km

t = s/v = 2.5/5

= 0.5 h

The time taken by man in returning home

t = s/v_{1} = 2.5/7.5

= 0.333h.

(i) Time travel 0 to 30 minutes

**Average velocity:** 𝚫x/ 𝚫t

= 𝚫x/𝚫t = 2.5/0.5

= 5 km h^{-1}

(ii) Average velocity (0 – 50 min) = 𝚫x/𝚫t = 2.5 + 2.5 / 0.5 + 0.333

= 5/ 0.833

= 8 km h^{-1}

(iii) Average velocity (0 – 40 min) = 𝚫x/𝚫t

= 1.8751 km h^{-1}

Average speed (0 – 40)

= Total distance / Total time

= 5.625 km h^{-1}.

**11. In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?**

Ans:

During very small intervals of time (𝚫t → 0), the direction of the motion of the body is not supposed to change, hence there is no difference between the total path length (distance) and the magnitude of displacement. Hence instantaneous speed is always equal to the magnitude of instantaneous velocity.

**12. Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.**

**(a) **

Ans: This graph does not represent one dimensional motion because A particle cannot have two different positions at the same time.

**(b) **

Ans: This graph does not represent one dimensional motion because, This is because at a particular time t, the particle has two motions in opposite direction.

**(c) **

Ans: This graph does not represent one dimensional motion because, this graph tells that the practical can have the negative speed but, the speed of the particle can never be negative.

**(d) **

Ans: This graph does not represent one dimensional motion because, Total path length of a particle can never decrease with time.

**13. Figure 2.11shows the x-t plot of one dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.**

Ans: It is not correct to say that the practical moves in a straight line for t<0 and on a parabolic path for t>0 because the graph can not show the path of the particle. This represents uniform motion, meaning the particle is moving in a straight line with a constant velocity.

**14. A police van moving on a highway with a speed of 30 km h**^{–1}** fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h**^{–1}**. If the muzzle speed of the bullet is 150 m s**^{–1}**, at what speed does the bullet hit the thief’s car? (Note: Obtain that speed which is relevant for damaging the thief’s car).**

Ans: **Speed of the van:** v_{van} = 30 km/h

**Speed of the thief’s car:** v_{thief} = 192 km/h

Muzzle speed of the bullet is = 150 m/s

**Now, we need to convert the speed from km/h to m/s:**

(i) v_{van} = 30 km/h = 1000 m / 1 km x 1 h / 3600 s

= 8.33m/s.

(ii) v_{thief} = 192 km/h = 1000m / 1 km x 1 h / 3600 s

= 53.33 m/s.

**The sum of the police van’s speed and the muzzle speed of the bullet:**

= v_{van} + v_{bullet} = 8.33 m/s + 150 m/s = 158.33 m/s

**The relative speed of the bullet to the thief’s car:**

v_{relative} = v_{bullet} – v_{thief} = 158.33 m/s – 53.33 m/s

= 105m/s.

**15. Suggest a suitable physical situation for each of the following graphs (Fig 2.12):**

**(a) **

Ans: The graph shows that initially the body was at rest. Then gains constant speed for some time.A ball at rest on a smooth floor is kicked, it rebounds from a wall with reduced speed and moves to the opposite wall which stops it.

**(b) **

Ans: The graph shows that the velocity decreases from a positive value to a negative value and at an instant its value decreases suddenly. Hence, it represent a physical situation such as freely falling ball on striking the ground rebounds with reduced speed after each hit against the ground.

**(c)**

Ans: The graph shows that the body gets accelerated during a short duration only. Hence it represents a physical situation such as a ball moving with a uniform speed is hit with a bat for a very short time-interval.

**16. Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.**

Ans: In the same S.H.M. acceleration,a = – ω^{2}x, Where ω is constant.

(i) At time t = 0.3s, x is negative.

The slope of x – t is plot negative.

Therefore position and velocity in negative.

a = – ω2x acceleration is positive.

(ii) At time t = 1.2s, x is positive, the slope of x-t plot is positive.

Hence the position and velocity are also positive.

Since, – ω^{2}x, acceleration is negative.

(iii) t = − 1.2 s In this time interval, x is negative.

The slope of the x-t plot is positive.

Hence, the velocity comes to be positive.

Finally acceleration ‘a’ is also positive.

**17. Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.**

Ans: The average speed is the ratio of the total distance travelled to the total time taken during the motion. It is the slope of the x-t plot shown in the figure. It is clear from the graph that the magnitude of slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. Moreover, the slope is positive in interval (1) and interval (2) but it is negative in the interval (3). Hence, the average velocity is positive in interval (1) and (2) but negative in interval (3).

**18. Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?**

Ans: The slope of the speed-time graph of the particle gives information about the acceleration of the particle. The magnitude of the average acceleration of the particle depends upon the slope of the graph in the particular interval.

(i) The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

(ii) It is clear from the figure that average speed is greatest in the interval (3) and the least in the interval (1).

(iii) The speed v is positive in all the three intervals. Further, in interval (1), the slope of 2 (t) graph is positive, while in interval (2), the slope is negative. Because, in interval (3), (v-t) graph is parallel to the time-axis, hence, a is zero in this interval.

(iv) Tangents to points A, B, C, and D are all parallel to the time axis. Hence, the slope is zero at these points. Therefore, the acceleration of the particle at these instants will also be zero.