NCERT Solutions Class 11 Physics Chapter 2 - Motion in a Straight Line

Q2.1: 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.
Ans: (a), (b)
Explanation:
(a) The length of a railway carriage is very small compared with the distance between two stations. For describing its motion from one station to the next, all parts of the carriage move nearly the same way. Therefore it can be treated as a point object.
(b) The size of the monkey is very small compared with the radius (or overall size) of the circular track. For the purpose of describing the monkey's motion around the track, its finite size does not affect the position of its centre and it can be treated approximately as a point.
(c) A spinning cricket ball undergoes important rotational motion and its size is comparable with the distance over which it changes direction on bouncing. Its rotation and finite radius matter, so it cannot be treated as a point for that motion.
(d) A tumbling beaker has significant rotation and changes of orientation while it falls; its shape and size affect its motion. Hence it cannot be considered a point object.

Q2.2:

Fig 2.9

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. 2.19. 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).
Ans: (a) A lives closer to school than B.
(b) A starts from school earlier than B.
(c) B walks faster than A.
(d) A and B reach home at the same time.
(e) B overtakes A once on the road.
Explanation:
(a) On the x-t graph the final x-value for A (location P) is smaller than that for B (location Q). Hence OP < OQ and A lives closer to the school than B.
(b) At x = 0 (the school) A has t = 0 while B has a positive starting time. Thus A started earlier.
(c) Speed is given by the slope of the x-t graph. The slope for B is larger than that for A, so B walks faster.
(d) Both plots reach their respective final positions at the same value of t on the time axis; therefore they reach home at the same time.
(e) B starts later but moves faster. The two x-t lines cross once, so B overtakes A once on the road.

Q2.3: 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: Speed while walking = 5 km/h.
Distance home ↔ office = 2.5 km.
Calculation and timings:
• Time to walk from home to office = distance / speed = 2.5 / 5 = 0.5 h = 30 min. So she reaches office at 9:30 am.
• She stays at office from 9:30 am to 5:00 pm (rest of day).
• Return by auto: speed = 25 km/h, time = 2.5 / 25 = 0.1 h = 6 min. So she leaves at 5:00 pm and arrives home at 5:06 pm.
Plot instructions (choose convenient scales):
• Horizontal axis: time from 9:00 am to 5:10 pm; mark hours and minutes suitably.
• Vertical axis: position x measured from home (x = 0 at home, x = 2.5 km at office).
• From 9:00 to 9:30 draw a straight line of slope 5 km/h rising from 0 to 2.5 km.
• From 9:30 to 17:00 draw a horizontal line at x = 2.5 km (rest at office).
• From 17:00 to 17:06 draw a steep straight line descending from 2.5 km to 0 with slope -25 km/h.
The required x-t graph is shown in the figure.  

Q2.4: 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: Each step length = 1 m, time per step = 1 s.
One cycle: 5 steps forward and 3 steps backward → net forward displacement per cycle = 5 - 3 = 2 m.
Time per cycle = 5 + 3 = 8 s.
Distance vs time (cycles):
• After 1 cycle (8 s): 2 m.
• After 2 cycles (16 s): 4 m.
• After 3 cycles (24 s): 6 m.
• After 4 cycles (32 s): 8 m.
Now next forward run of 5 steps begins. From 8 m he moves forward step by step: after 1st step (33 s) → 9 m; 2nd (34 s) → 10 m; 3rd (35 s) → 11 m; 4th (36 s) → 12 m; 5th (37 s) → 13 m and falls into the pit.
Therefore, total time to reach 13 m = 37 s.
Graph: The x-t graph is a staircase-like plot rising in short straight segments for forward steps and dropping slightly for backward steps, with an overall upward trend. See figure placeholder .

Q2.5: A car moving along a straight highway with a 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: Given:
Initial speed u = 126 km/h = 126 × (1000/3600) = 35 m/s.
Final speed v = 0.
Stopping distance s = 200 m.
Retardation (a):
Retardation magnitude = 3.0625 m/s² (negative sign shows deceleration).
Time to stop (t):
Use v = u + a tAnswer: Retardation = -3.06 m/s² (approximately), time to stop ≈ 11.43 s.

Q2.6: 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?
(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).
Ans: (a) The acceleration is downward (i.e., directed towards the centre of the Earth).
(b) At the highest point velocity = 0, acceleration = 9.8 m/s² downward.
(c) With the chosen sign convention (x = 0 at highest point, downward positive):
• Position x > 0 during both upward and downward motion (because the ball is below the highest point).
• Velocity v < 0 during the upward motion (the ball moves upward, which is negative in this convention); v > 0 during the downward motion.
• Acceleration a > 0 throughout the motion (downward and equal to +g in this convention).
(d) Given u = 29.4 m/s and g = 9.8 m/s².

From first equation of motion, time of ascent (t) is given as:

Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.
Answer: Height = 44.1 m, total time = 6.0 s.

Q2.7: 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
(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.
Ans: (a) True
(b) False
(c) True
(d) False
Explanation:
(a) True. Example: A ball thrown vertically upwards has speed zero at the highest point, but acceleration is non-zero (g downward). Thus zero speed does not imply zero acceleration.
(b) False. Speed is the magnitude of velocity. If speed = 0 then the magnitude of velocity is zero and so the velocity vector is zero.
(c) True (in one-dimensional motion). Constant speed in one dimension implies constant velocity (no change in direction), therefore acceleration (rate of change of velocity) is zero. Example: a car cruising at steady speed on a straight, level road.
(d) False in general. Positive acceleration means velocity is changing in the positive direction. Whether the speed increases depends on the sign of velocity. If velocity is already positive, positive acceleration speeds the particle up. But if velocity is negative (motion in the negative direction) and acceleration is positive, the speed decreases until velocity becomes zero; afterwards the object speeds up in the positive direction. Example: a body thrown upwards has negative velocity (upwards if downward is positive) while acceleration due to gravity is positive in the chosen sign convention: the body slows down while v and a have opposite signs.

Q2.8: 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: Given: initial height h = 90 m, initial speed u = 0, g = 9.8 m/s².
Time to first impact:
Speed just before first impact: 
v = g t ≈ 9.8 × 4.29
 ≈ 42.04 m/s.
After collision the speed is reduced by one tenth, so rebound speed 

u_r =
Time to reach the next maximum height after rebound: 
t′ = u_r / g =So the ball returns to the floor for the second impact at t = 4.29 + 3.86 ≈ 8.15 s. 
The speed just before the second impact equals the rebound speed from the first impact (by symmetry) ≈ 37.84 m/s. 
After the second impact speed becomesSpeed-time graph shape (0 to 12 s):
• From t = 0 to 4.29 s the speed increases linearly from 0 to ≈ 42.04 m/s (slope = g).
• At t ≈ 4.29 s there is an instantaneous drop to 0.9 × 42.04 ≈ 37.84 m/s (rebound).
• From 4.29 to 8.15 s the speed decreases linearly to zero (at the top of rebound) and then increases again to ≈ 37.84 m/s at impact - when plotted as speed (non-negative) this segment appears as two straight lines joining at the top; however since speed is always non-negative it is the absolute value of velocity.
• At t ≈ 8.15 s the ball rebounds again and speed is reduced to ~34.05 m/s; between 8.15 s and 12 s the speed varies linearly as the ball rises and falls for the second rebound (partially shown).
The piecewise linear segments and sudden drops at impacts produce the speed-time graph shown in the figure. See placeholders -.

Q2.9: 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;
(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 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, suppose a particle moves from point A to point B and then, comes back to a point, C taking a total time t, as shown below. Then, the magnitude of displacement of the particle = AC.

Whereas, total path length = AB + BC
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
(b) 


Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.

Q2.10: 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: Time taken by the man to reach the market from home,
Time taken by the man to reach home from the market,
Total time taken in the whole journey = 30 + 20 = 50 min


Time = 50 min = 
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km


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
=
Net displacement = 2.5 – 1.25 = 1.25 km
Total distance travelled = 2.5 +1.25 = 3.75 km

Q2.11: 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: Instantaneous velocity is defined as the derivative dx/dt in the limit dt → 0. Over an infinitesimally small time dt the particle cannot reverse direction; the path length in that interval equals the magnitude of the infinitesimal displacement. Therefore instantaneous speed (the limit of path length per dt) equals the magnitude of instantaneous velocity |dx/dt|. Hence instantaneous speed = magnitude of instantaneous velocity.

Q2.12: Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
(a)

(b)

(c)

(d)

Ans: (a), (b), (c), (d) - each has a reason why it cannot represent one-dimensional motion:

(a) The x-t graph shows two different x values at the same time (a vertical line). That is impossible: a particle cannot be at two positions simultaneously in one-dimensional motion.

(b) The v-t graph shows two different velocities at the same instant (vertical overlap). A particle cannot have two distinct velocities at the same time.

(c) The v-t graph takes negative values for speed. Speed is a scalar and cannot be negative. If the plot is intended to be velocity then negative values are allowed, but the question labels it as speed, so it is invalid.

(d) The v-t graph implies the total path length decreases with time (area under speed-time curve reduces). Path length cannot decrease with time, so this plot cannot represent speed vs time for a particle.

Q2.13: Figure 2.11 shows the x-tplot 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.

(Fig 2.11)

Ans: No

The x-t graph only gives the coordinate x as a function of time. It does not show the geometric trajectory in space (which for one-dimensional motion is always a straight line). Saying the particle moves on a parabolic path for t > 0 confuses the x-t plot with the spatial path. A suitable physical context for the given x-t plot is: a body is held at some height for t < 0 (no change in x), and at t = 0 it is released and then undergoes uniformly accelerated motion (free fall) for t > 0. In other words, it resembles a freely falling body that was initially held at rest and then released at t = 0.

Q2.14: 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, with what speed does the bullet hit the thief's car ? (Note: Obtain that speed which is relevant for damaging the thief's car).
Ans: Convert speeds to m/s:
v_police = 30 km/h = 30 × 1000 / 3600 = 8.33 m/s.
v_thief = 192 km/h = 192 × 1000 / 3600 = 53.33 m/s.
Muzzle speed relative to van = 150 m/s. Speed of bullet relative to ground = 
150 + 8.33 = 158.33 m/s (same direction as motion).
Speed at which bullet approaches the thief's car = bullet speed relative to ground - thief car speed =
 158.33 - 53.33 = 105.00 m/s.
Answer: The bullet strikes the thief's car with speed ≈ 105 m/s relative to the car.

Q2.15: Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
(a)

(b)

(c)

(Fig: 3.22)

Ans: (a) The x-t graph shows an object initially at rest, then accelerating so that speed increases to some constant value, then slowing to rest, and then reversing direction and acquiring a constant speed in the opposite direction. A physical situation: a football resting is kicked and moves forward, bounces back from a wall (or is returned by another player) and then eventually comes to rest after moving back.

(b) The v-t graph shows the magnitude of velocity decreasing and changing sign on rebound, typical of a ball dropped on a hard floor: it strikes, rebounds with reduced speed (each rebound giving a smaller peak), and eventually comes to rest.

(c) The a-t graph shows acceleration non-zero for a short interval and zero otherwise. Example: A hammer moving with uniform velocity strikes a nail; during the short impact time acceleration changes (force on nail), then returns to zero as motion again becomes uniform.

Q2.16: 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 Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

(Fig: 2.13)

Ans: At t = 0.3 s: position x is negative, velocity v is negative, acceleration a is positive.
At t = 1.2 s: position x is positive, velocity v is positive, acceleration a is negative.
At t = -1.2 s: position x is negative, velocity v is positive, acceleration a is positive.
Reason: For SHM a = -ω² x. 
The sign of acceleration is opposite to the sign of x. 
The sign of velocity follows from the slope of the x-t plot: if x is decreasing with time at the chosen instant, v is negative; if x is increasing, v is positive.
 These sign assignments are consistent with the graph segments shown.

Q2.17: Figure 2.14 gives the x-tplot 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.

(Fig: 2.14)

Ans: Average speed is greatest in interval 3 and least in interval 2.
Sign of average velocity: intervals 1 and 2 → positive; interval 3 → negative.
Explanation: Average speed over an interval equals the magnitude of the slope of the x-t curve in that interval. The slope is largest (in magnitude) in interval 3, so average speed is greatest there. The slope is smallest in interval 2, so average speed is least. The slope is positive in intervals 1 and 2 (hence average velocity positive) and negative in interval 3 (hence average velocity negative).

Q2.18: 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?

(Fig: 2.15)

Ans: Average acceleration is greatest in interval 2.
Average speed is greatest in interval 3.
Signs: v is positive in intervals 1, 2 and 3 (speed is non-negative and the motion is in the chosen positive direction). Acceleration a is positive in intervals 1 and 3 (sloping upward or zero) and negative in interval 2 (sloping downward).
Accelerations at points A, B, C and D are zero.
Explanation: Acceleration equals the slope of the v-t graph. The slope magnitude is largest (negative but large in magnitude) in interval 2, so average acceleration magnitude is greatest there. The height of the curve above the time axis gives speed; it is largest in interval 3. Points A-D lie on horizontal segments of the graph so slope = 0; hence acceleration = 0 at those points.

Old NCERT Solutions

Q2: Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s². If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them
Ans: Convert speed: 
72 km/h = 20 m/s. 
Time t = 50 s.
Distance covered by A in 50 s: 
s_A = 20 × 50 = 1000 m.
For train B (initial speed 20 m/s, acceleration a = 1 m/s²):
= 1000 + 1250 = 2250 m.
At the instant given the guard of B just brushes the driver of A. 
To clear the driver of A, B must cover its initial lead plus the lengths of both trains. Let d be the original distance between driver of A and guard of B. T
hen during 50 s B covers distance equal to d + length(A) + length(B) + distance covered by A in the same time.
 Rearranging gives
d = s_B - s_A - (length A + length B) 
= 2250 - 1000 - 800 = 450 m.
Answer: The original distance between them was 450 m.

Q3: 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?
Ans: Convert speeds: 
v_A = 36 km/h = 10 m/s, 
v_B = v_C = 54 km/h = 15 m/s.
Relative speed of C towards A = 15 + 10 = 25 m/s. 
Distance AC = 1000 m 
 time for C to reach A = 1000 / 25 = 40 s. 
Thus B must cover its 1000 m to the overtaking point in ≤ 40 s.
If B accelerates from initial 15 m/s with acceleration a and must cover 1000 m in t = 40 s:
Use Answer: Minimum acceleration required by B is 0.5 m/s².

Q4: Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. 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?
Ans: Let speed of bus = V km/h, cyclist speed v = 20 km/h.
Relative speed of buses moving in same direction as cyclist = V - 20 km/h
The bus went past the cyclist every 18 min i.e., (when he moves in the direction of the bus).
Distance covered by the bus = … (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to
Both equations (i) and (ii) are equal.

Relative speed of the bus moving in the opposite direction of the cyclist= (V + 20) km/h
Time taken by the bus to go past the cyclist

From equations (iii) and (iv), we get

Substituting the value of V in equation (iv), we get

Q5: A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n^th second (n = 1,2,3....) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Ans: Distance covered by a body in n^th second is given by the relation

Where,
u = Initial velocity
a = Acceleration
n = Time = 1, 2, 3, ..... ,n
In the given case,
u = 0 and a = 1 m/s²

This relation shows that:
D_n ∝ n … (iii)
Now, substituting different values of n in equation (iii), we get the following table:

The plot between n and D_n will be a straight line as shown:

.

Q6: A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Ans: Case I (lift stationary): 
u = 49 m/s, 
g = 9.8 m/s² (upward positive). 
From first equation of motion, time of ascent (t) is given as:
Total time up and down = 2 × 5 = 10 s.
Case II (lift moving up at 5 m/s): 
The boy and lift share the same upward velocity; 
the ball's speed relative to the boy at the moment of release is still 49 m/s (given as the maximum he can impart). 
The relative motion of ball and boy is the same as in case I, so the round-trip time measured by the boy is still 10 s.
Answer: Time to return in both cases = 10 s.

Q7: On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h^-1 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^-1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt ?
(b) speed of the child running opposite to the direction of motion of the belt ?
(c) time taken by the child in (a) and (b) ?
Which of the answers alter if motion is viewed by one of the parents?

(Fig: 3.26)

Ans: Convert speeds to consistent units or use km/h then convert final times to seconds.
Child's speed relative to belt = 9 km/h; 
belt speed = 4 km/h.
(a) When child runs in same direction as belt, speed relative to ground
v_bB = v_b +v_B = 9 + 4 = 13 km/h 
(b) When child runs opposite belt's motion, speed relative to ground
v_bB = v_b +(– v_B) = 9 - 4 = 5 km/h.
(c) Distance between parents on the belt = 50 m. Parents and child are on the belt, so their relative speed (child relative to a parent) is 9 km/h = 2.5 m/s. Time to cover 50 m (as seen by parents) = 50 / 2.5 = 20 s.
Which answers change for a parent observer?
For an observer on the belt (a parent), the child's speed is 9 km/h in both directions (by definition of the child's speed with respect to the belt); the times measured by parents use 9 km/h (2.5 m/s) and so are different from the times an outside stationary observer would compute using 13 km/h or 5 km/h. Thus the numerical values in (a) and (b) are specific to the stationary observer; for a parent the child's speed is always 9 km/h and the time between parents remains 20 s.

Q8: Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m/s and 30 m/s. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g= 10 m/s². Give the equations for the linear and curved parts of the plot.

Ans: For first stone:
Initial velocity, u_I = 15 m/s
Acceleration, a = –g = – 10 m/s²
Using the relation,

When this stone hits the ground, x₁ = 0
∴– 5t²  15t 200 = 0
t² – 3t – 40 = 0
t² – 8t+ 5t – 40 = 0
t (t – 8)+ 5 (t – 8) = 0
t = 8 s or t = – 5 s
Since the stone was projected at time t = 0, the negative sign before time is meaningless.
∴t = 8 s
For second stone:
Initial velocity, u_II = 30 m/s
Acceleration, a = –g = – 10 m/s²
Using the relation,

At the moment when this stone hits the ground; x₂ = 0
– 5t² 30 t 200 = 0
t² – 6t – 40 = 0
t² – 10t+ 4t 40 = 0
t (t – 10) +4 (t – 10) = 0
(t – 10) (t+ 4) = 0
t = 10 s or t = – 4 s
Here again, the negative sign is meaningless.
∴t = 10 s
Subtracting equations (i) and (ii), we get

Equation (iii) represents the linear path of both stones. Due to this linear relation between (x₂ – x₁) and t, the path remains a straight line till 8 s.
Maximum separation between the two stones is at t = 8 s.
(x₂ – x₁)_max = 15× 8 = 120 m
This is in accordance with the given graph.
After 8 s, only second stone is in motion whose variation with time is given by the quadratic equation:
x₂ – x₁ = 200 +30t – 5t²
Hence, the equation of linear and curved path is given by
x₂ – x₁ = 15t (Linear path)
x₂ ­– x₁ = 200 +30t – 5t² (Curved path)

Q9: The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

(Fig. 3.28)

What is the average speed of the particle over the intervals in (a) and (b)?

Ans: (a) Distance = area under the speed-time graph from t = 0 to 10 s.
Average speed =  and .

(b)  Let s₁ and s₂ be the distances covered by the particle between time
t = 2 s to 5 s and t = 5 s to 6 s respectively.
Total distance (s) covered by the particle in time t = 2 s to 6 s
s = s₁ + s₂ … (i)
For distance s₁:
Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.
Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:
v = u + at
Where,
v = Final velocity of the particle
12 = 0 a′ × 5

Again, from first equation of motion, we have
v = u + at
= 0 2.4 × 2 = 4.8 m/s
Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

For distance s₂:
Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.
From first equation of motion,
v = u  at (where v = 0 as the particle finally comes to rest)
0 = 12 a″ × 5

Distance travelled by the particle in s1 (i.e., between t = 5 s and t = 6 s)

From equations (i), (ii), and (iii), we get
s = 25.2 + 10.8 = 36 m

Ques 10: The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29:

Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂?
(a) x(t₂) = x(t₁) + v(t₁)(t₂-t₁) + 1/2 a (t₂-t₁)²
(b) v(t₂) = v(t₁) + a (t₂-t₁)
(c) v_Average = (x(t₂) - x(t₁)) / (t₂ - t₁)
(d) a_Average = (v(t₂) - v(t₁)) / (t₂ - t₁)
(e) x(t₂) = x(t₁) + v_Average(t₂ - t₁) + 1/2 a_Average(t₂ - t₁)²
(f) x(t₂) - x(t₁) = area under the v-t curve bounded by the t-axis and the dotted line shown.
Ans: The correct formulae are (c), (d) and (f).
Explanation: (c) and (d) are definitions of average velocity and average acceleration respectively and hold generally. (f) is correct because displacement equals the integral (area under the v-t curve) between the times. Formulae (a), (b) and (e) assume a single constant acceleration a over the interval; but the given v-t graph shows non-uniform slope (acceleration is not constant), so those forms do not apply over the whole interval.