Motion In a Straight Line
Based on NCERT Class 11th Physics Chapter 3
Exercise
Q 3.1: 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.
b) A monkey sitting on top of a man cycling smoothly on a circular track.
c) A spinning cricket ball that turns sharply on hitting the ground.
d) A tumbling beaker that has slipped off the edge of a table.
Answer: (b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track. (c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object. (d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
Q 3.2 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. Choose the correct
entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
Answer: Explanation: A lives closer to school than B : In the given x–t graph, it can be observed that distance OP < OQ. Hence, the distance of school from the A’s home is less than that from B’s home. A starts from school earlier than B :In the given graph, it can be observed that for x = 0, t = 0 for A, whereas for x = 0, t has some finite value for B. Thus, A starts his journey from school earlier than B. B walks faster than A : In the given x–t graph, it can be observed that the slope of B is greater than that of A. Since the slope of the x–t graph gives the speed, a greater slope means that the speed of B is greater than the speed A. B overtakes A once on the road : B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.
A and B reach home at the same time : It is clear from the given graph that both A and B reach their respective homes at the same time.
Q 3. 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.
Answer: Distance covered while walking = 2.5 km. If O is regarded as the origin for both time and distance, then at t = 9.00 am, x = 0 OA is the x-t graph of the motion when the woman walks from her home to office. Her stay in the office from 9.30 am to 5.00 pm is represented, by the straight line AB in the graph. Now, time taken to return home by an auto = 2.5/25 h =1/10 h =6 minute So, at t = 5.06 pm, x = 0 This motion is represented by the straight line BC in the graph. While drawing the x-t graph, the scales chosen are as under:
Speed while walking = 5 km/h
Time taken to reach office while walking = (2.5/5 ) h=1/2 h
and at t = 9.30 am, x = 2.5 km
Along time-axis, one division equals 1 hour.
Along positive-axis, one division equals 0.5 km.
Q3.4 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.
Answer: Distance covered with 1 step = 1 m Time taken = 1 s Drunkard covers 2 m in 8 s. After covered of distance 8m, in the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit ( he will not move backward 3 steps) .
Time taken to move first 5 m forward = 5 s
Time taken to move 3 m backward = 3 s
Net distance covered = 5 – 3 = 2 m
Net time taken to cover 2 m = 8 s
Drunkard covered 4 m in 16 s.
Drunkard covered 6 m in 24 s.
Drunkard covered 8 m in 32 s.
Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s The x-t graph of the drunkard’s motion can be shown as:
Q 3. 5. A jet airplane travelling at the speed of 500 km h-1 ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer: Speed of the jet airplane, \(v_{jet}\) = 500 km/h Relative speed of its products of combustion with respect to the plane, Speed of its products of combustion with respect to the ground = \(v’_{smoke}\) Relative speed of its products of combustion with respect to the airplane, – 1500 = \(v’_{smoke}\) – 500 \(v’_{smoke}\) = – 1000 km/h The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane
\(v_{smoke}\) = – 1500 km/h
\(v_{smoke}\) =\(v’_{smoke}\) – \(v_{jet}\)
Q 3. 6. 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?
Answer: Initial velocity of the car, \(u\) = 126 km/h = 35 m/s \(v^2-u^2\)=\(2as\) \((0)^2-(35)^2\) =\(2× a × 200\) \(a\) = \(- \frac{35× 35}{2 × 200}\) = – 3.06 m/s2
Final velocity of the car, \(v\) = 0
Distance covered by the car before coming to rest, \(s\) = 200 m
Retardation produced in the car = \(a\)
From third equation of motion, \(a\) can be calculated as:
Q 3. 7. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 71 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms-1. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Answer: For train A: \(s_1\) = \(ut+\frac{1}{2}a_1t^2\) = 20 × 50 + 0 = 1000 m For train B: \(s_2\) = \(ut+\frac{1}{2}at^2\) = \(20 × 50 +\frac{1}{2} × 1 × 50^2\) =2250m Hence, the original distance between the driver of train A and the guard of train B
Initial velocity, \(u\) = 72 km/h = 20 m/s
Time, \(t\) = 50 s
Acceleration, \(a_1\) = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance \(s_1\) covered by train A can be obtained as:
Initial velocity, \(u\) = 72 km/h = 20 m/s
Time, \(t\) = 50 s
Acceleration, \(a\) = 1 m/s2
From second equation of motion, distance \(s_2\) covered by train B can be obtained as:
= 2250 –1000 = 1250 m.
Q 3. 8. On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer: Velocity of car A, \(v_A\) = 36 km/h = 10 m/s Relative velocity of car B with respect to car A, Relative velocity of car C with respect to car A, At a certain instance, both cars B and C are at the same distance from car A i.e., Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as: \(s\) = \(ut+\frac{1}{2}at^2\) 1000 = \(5 × 40 +\frac{1}{2} × a × 40^2\) \(a\) = \(\frac{1600}{1600}\) = 1 m/s2
Velocity of car B, \(v_B\) = 54 km/h = 15 m/s
Velocity of car C, \(v_C\) = 54 km/h = 15 m/s
\(v_{BA}\) = \(v_B – v_A\) = 15 – 10 = 5 m/s
\(v_{CA}\) = \(v_C – (- v_A)\) = 15 + 10 = 25 m/s
s = 1 km = 1000 m
Time taken (t) by car C to cover 1000 m = 1000/25=40 s
Q 3. 9. Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minute. A man cycling with a speed of 20 km h-1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer: Let V be the speed of the bus running between towns A and B. Relative speed of the bus moving in the direction of the cyclist The bus went past the cyclist every 18 min i.e., \(\frac{18}{60}\) h (when he moves in the direction of the bus). Distance covered by the bus = \(v_b – 20 \frac{18}{60}\) ………….. (i) Since one bus leaves after every T minutes, the distance travelled by the bus will be equal \(v_b × \frac{T}{60}\) ………….. (ii) Both equations (i) and (ii) are equal. \((v_b – 20) \frac{18}{60}\) = \(v_b × \frac{T}{60}\) ………….. (iii) Relative speed of the bus moving in the opposite direction of the cyclist Time taken by the bus to go past the cyclist = 6 min = \(\frac{6}{60}\) h \((v_b+ 20)\) \(\frac{6}{60}\) = \(v_b × \frac{T}{60}\) ………….. (iv) From equations (iii) and (iv), we get \((v_b+ 20)\) \(\frac{6}{60}\) = \((v_b – 20) \frac{18}{60}\) \(v_b+ 20\) = \(3v_b – 60\) \(v_b\) = 40 km/h Substituting the value of \(v_b\) in equation (iv), we get \((40+ 20)\) \(\frac{6}{60}\) = \(40 × \frac{T}{60}\) \(T\) = \(\frac{360}{40}\) = 9 min
Speed of the cyclist, \(v_c\) = 20 km/h
= \(v_b – v_c\) = \(v_b – 20\) km/h
to:
= \((v_b+ 20)\) km/h
Q3. 10. A player throws a ball upwards with an initial speed of 29.4 ms-1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(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.
(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).
Answer: (a) The direction of acceleration during the upward motion of the ball is downward: Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth. (b) Velocity = 0, acceleration = 9.8 m/s2 At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8 m/s2. (c) x > 0 for both up and down motions v < 0 for up and v > 0 for down motion: During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. (d) 44.1 m, 6 s : Initial velocity of the ball, u = 29.4 m/s From third equation of motion, height (s) can be calculated as: \(v^2-u^2\) = \(2gs\) \(s\) = \(v^2-u^2 \over 2g\) = \(0^2 – 29.4^2 \over {2 × -9.8}\) = 44.1 m From first equation of motion, time of ascent (t) is given as: \(v\) = \(u + at\) \(t\)= \(\frac{v-u}{a}\) = \(\frac{0 – 29.4}{-9.8}\) = 3 s Time of ascent = Time of descent
During downward motion, the signs of position, velocity, and acceleration are all positive. a > 0 throughout the motion
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a = – g = – 9.8 m/s2
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.
Q 3. 11. 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.
(b) with, zero speed may have non-zero velocity.
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer: a) True : When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point. b) False : Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero. c) True : A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero. d) False : This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards. This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.
Q3.12 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.
Answer: Ball is dropped from a height, \(s\) = 90 m \(s\) = \(ut+\frac{1}{2}at^2\) 90 = \(0+\frac{1}{2}9.8 t^2\) \(t\) =\(\sqrt 18.38\) = 4.29 s From first equation of motion, final velocity is given as: \(v\) =\( u + at\) Rebound velocity of the ball, \(u_r\) = \(\frac{9}{10}v\) …..(the ball loses one tenth of its speed) = \(\frac{9}{10} × 42.04\) = 37.84 m/s Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as: \(v\) =\( ur + at′\) t’=3.86 s Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s
Initial velocity of the ball, \(u\) = 0
Acceleration, \(a\) = \(g\) = 9.8 m/s2
Final velocity of the ball = \(v\)
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
= 0 + 9.8 × 4.29 = 42.04 m/s
0 = 37.84 + (– 9.8) t’
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor = = \(\frac{9}{10} × 37.84 \) = 34.05 m/s
The speed-time graph of the ball is represented in the given figure as: 
Q3.13 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;
(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].
Answer: (a) Suppose a particle goes from point A to B along a straight path and returns to A along the same path. The magnitude of the displacement of the particle is zero, because the particle has returned to its initial position. The total length of path covered by the particle is AB + BA = AB + AB = 2 AB. Thus, the second quantity is greater than the first, (b) Suppose, in the above example, the particle takes time t to cover the whole journey. Then, the magnitude of the average velocity of the particle over time-interval t is = Magnitude of displacement /Time-interval =0/t =0 While the average speed of the particle over the same time-interval is =Total path length /Time-interval= 2 AB /t Again, the second quantity (average speed) is greater than the first (magnitude of average velocity). In both the above cases, the two quantities are equal if the particle moves from one point to another along a straight path in the same direction only.
Q3.14. 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!]
Answer: Time taken by the man to reach the market from home –2.5 km away with a speed of 5 km h-1 \(t_1\) = \(\frac{2.5}{5}\) = \(\frac{1}{2}\) h = 30 min Time taken by the man to reach home from the market – 2.5 km with a speed of 7.5 km h-1 \(t_2\) = \(\frac{2.5}{7.5}\) = \(\frac{1}{3}\) h = 20 min Total time taken in the whole journey = 30 + 20 = 50 min Average Velocity = \(\frac{Displacement}{Time}\) = \(\frac{2.5}{1/2}\) = 5Km/h ….. (a-i) Average Speed = \(\frac{Distance}{Time}\) = \(\frac{2.5}{1/2}\) = 5Km/h ….. (b-i) Time = 50 min = \(\frac{5}{6}\) h Net displacement = 0 Average Velocity = \(\frac{Displacement}{Time}\) = \(\frac{0}{5/6}\) = 0 ….. (a-ii) Average Speed = \(\frac{Distance}{Time}\) = \(\frac{5}{5/6}\) = 6 Km/h ….. (b-ii) Speed of the man = 7.5 km Distance travelled in first 30 min = 2.5 km Distance travelled by the man (from market to home) in the next 10 min = \(7.5 ×\frac{10}{60}\) = 1.25 Km Net displacement = 2.5 – 1.25 = 1.25 km Average Velocity = \(\frac{Displacement}{Time}\) = \(\frac{1.25}{40/60}\) = 1.875 km/h ….. (a-iii) Average Speed = \(\frac{Distance}{Time}\) = \(\frac{3.75}{40/60}\) = 5.625 Km/h ….. (b-iii)
Total distance = 2.5 + 2.5 = 5 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Q3.15 In Exercises 3.13 and 3.14, 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 ?
Answer: Instantaneous velocity is the velocity of a particle at a particular instant of time. In this case of small interval of time, the magnitude of the displacement is effectively equal to the distance travelled by the particle in the same interval of time. Therefore, there is no distinction between instantaneous velocity and speed.
Q3.16 Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Answer: The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time. The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time. The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative. The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
Q3. 17. Figure shows 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.
Answer: It is not correct to say that the particle moves in a straight line for t < 0 (i.e., -ve) and on a parabolic path for t > 0 (i.e., + ve) because the x-t graph can not show the path of the particle. For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t =0.
Q 3. 18. 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 ms-1 , with what speed does the bullet hit the thief s car? (Note: Obtain that speed which is relevant for damaging the thief s car).
Answer: Speed of the police van, \(v_p\) = 30 km/h = 8.33 m/s Muzzle speed of the bullet, \(v_b\) = 150 m/s Speed of the thief’s car, \(v_t\) = 192 km/h = 53.33 m/s Since the bullet is fired from a moving van, its resultant speed can be obtained as: Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as: \(v_{bt}\) = \(v_b – v_t\) = 158.33 – 53.33 = 105 m/s
= 150 + 8.33 = 158.33 m/s
Q3. 19. Suggest a suitable physical situation for each of the following graphs:
Answer: (a) The given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime. (b) In the given v-t graph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero. (c) The given a-t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.
Q3. 20. Figure gives the x-t plot of a particle executing one¬dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Answer: For simple harmonic motion (SHM) of a particle, acceleration \(a\) is given by the relation: Negative, Negative, Positive (at t = 0.3 s) : t = 0.3 s In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive. Positive, Positive, Negative (at t = 1.2 s) :t = 1.2 s In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. Negative, Positive, Positive (at t = –1.2 s): t = – 1.2 s From equation (i), it can be inferred that the acceleration of the particle will be positive.
\(a\) = – ω2x ω → angular frequency …………..… (i)
However, using equation (i), acceleration of the particle comes to be negative.
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive.
Q3. 21. Figure 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.
Answer: Interval 3 (Greatest), Interval 2 (Least) Positive (Intervals 1 & 2), Negative (Interval 3) The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively.
Q3. 22. Figure 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?
Answer: Average acceleration is greatest in interval 2 Average speed is greatest in interval 3 v is positive in intervals 1, 2, and 3 a is positive in intervals 1 and 3 and negative in interval 2 a = 0 at A, B, C, D In interval 1: In interval 2: In interval 3: Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.
Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.