  # NCERT Solutions for Class 11 Physics Chapter 3 Motion in A Straight Line

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line is an essential tool that will help your class 11 preparation. They consist of answers to the questions given in the textbook together with important questions from CBSE previous year question papers and CBSE sample papers. Exemplar problems provide solutions that help you in your exam preparation and cross-check your preparation level. Ans.

(a) A lives closer to the school than B, because A has to cover shorter distances [OP < OQ],
(b) A starts from school earlier than B, because for x= 0, t = 0 for A but for B, t has some finite time.
(c) The slope of B is greater than that of A, therefore B walks faster than A.
(d) Both A and B will reach their home at the same time.
(e) At the point of intersection, B overtakes A on the roads once.

Q3. A woman starts from her home at 9.00 am, walks at a speed of 5 km/h 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 auto with a speed of 25 km/h. Choose suitable scales and
plot the x-t graph of her motion. Ans.

Distance till her office = 2.5 km.
Walking speed the woman= 5 km/h
Time taken to reach office while walking = (2.5/5 ) h=(1/2) h = 30 minutes

Speed of auto = 25 km/h

Time taken to reach home in auto = 2.5/25 = (1/10) h = 0.1 h = 6 minutes

In the graph, O is taken as the origin of the distance and the time, then at  t = 9.00 am, x = 0
and at t = 9.30 am, x = 2.5 km

OA is the portion on the x-t graph that represents her walk from home to the office. AB represents her time of stay in the office from 9.30 to 5. Her return journey is represented by BC.

Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards, followed again by 5 steps forward and 3 steps backwards, and so on. Each step is 1m 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 time taken to go one step is 1 second. In 5s he moves forward through a distance of 5m and then in next 3s he comes back by 3m. Therefore,  in 8s he covers 2m. So, to cover a distance of 8m he takes 32s. He must take another 5steps forward to fall into the pit. So, the total time taken is 32s + 5s = 37s to fall into a pit 13 m away.

Q.5. A jet airplane travelling at the speed of 500 km/h ejects its products of combustion at the speed of 1500 km/h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

Ans:
Speed of the jet airplane, VA= 500 km/h
Speed at which the combustion products are ejected relative to the jet plane, VB – VA
= – 1500 km/h
(The negative sign indicates that the combustion products move in a direction opposite to that of jet)
Speed of combustion products w.r.t. observer on the ground, VB – 500 = – 1500
VB = – 1500 + 500 = – 1000 km/h

Q6 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.

The initial velocity of the car = u

Final velocity of the car = v

Distance covered by the car before coming to rest = 200 m

Using the equation,

v = u + at

t = (v – u)/a = 11.44 sec.

Therefore, it takes 11.44 sec for the car to stop.

Q.7. 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 s–2. If after 50 s, the guard of B just
brushes past the driver of A, what was the original distance between them?

Ans:

Length of the train A and B = 400 m

Speed of both the trains = 72 km/h = 72 x (5/18) = 20m/s

Using the relation, s = ut + (1/2)at2

Distance covered by the train B

SB = uBt + (1/2)at2

Acceleration, a = 1 m/s

Time = 50 s

SB = (20 x 50) + (1/2) x 1 x (50)2

= 2250 m

Distance covered by the train A

SA = uAt + (1/2)at2

Acceleration, a = 0

SA = uAt  = 20 x 50 = 1000 m

Therefore, the original distance between the two trains = SB – SA = 2250 – 1000 = 1250 m

Q. 8. On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars B and
C approach car A in opposite directions with a speed of 54 km/h 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:

The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s

Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s

Speed of car C = – 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite direction)

Relative speed of A w.r.t C, VAC= VA – VB = 10 – (-15) = 25 m/s

Relative speed of B w.r.t A, VBA = VB – VA = 15 – 10 = 5 m/s

Distance between AB = Distance between AC = 1 km = 1000 m

Time taken by the car C to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 s

If a is the acceleration, then

s = ut + (1/2) at2

1000 = (5 x 40) + (1/2) a (40) 2

a = (1000 – 200)/ 800 = 1 m/s2

Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s2

Q. 9. Two towns A and B are connected by 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:

Speed of each bus = Vb

Speed of the cyclist = V= 20 km/h

The relative velocity of the buses plying in the direction of motion of cyclist is Vb – Vc
The buses playing in the direction of motion of the cyclist go past him after every 18 minutes i.e.(18/60) s.

Distance covered = (Vb – Vc ) x 18/60

Since the buses are leaving every T minutes. Therefore, the distance is equal to Vb x (T/60)

(Vb – Vc ) x 18/60 = Vb x (T/60) ——(1)

The relative velocity of the buses plying in the direction opposite to the motion of cyclist is Vb + Vc
The buses go past the cyclist every 6 minutes i.e.(6/60) s.

Distance covered = (Vb + Vc ) x 6/60

Therefore, (Vb +Vc ) x 6/60 = Vb x (T/60)——(2)

Dividing (2) by (1)

[(Vb – Vc ) x 18/60]/ [(Vb + Vc ) x 6/60 ]= [Vb x (T/60)] /[Vb x (T/60)]

(Vb – Vc ) 18/(Vb +Vc ) 6 = 1

(Vb – Vc )3 = (Vb +Vc )

Substituting  the value of Vc

(Vb – 20 )3= (Vb + 20 )

3Vb – 60 = Vb + 20

2Vb = 80

Vb = 80/2 = 40 km/h

To find the value of T, substitute the values of Vb and Vc in equation (1)

(Vb – Vc ) x 18/60 = Vb x (T/60)

(40 – 20) x (18/60) = 40 x (T/60)

T = (20 x 18) /40 = 9 minutes

Q.10. A player throws a ball upwards with an initial speed of 29.4 m/s.
(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 the 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 due to gravity always acts downwards towards the centre of the Earth.
(b) At the highest point of its motion the velocity of the ball will be zero but the acceleration due to gravity will be 9.8 m s–2  acting vertically downward.
(c) If we consider the highest point of ball motion as x = 0, t = 0 and vertically downward direction to be +ve direction of the x-axis, then
(i) During upward motion of the ball before reaching the highest point position ,x = +ve, velocity, v = -ve and acceleration, a =  +ve.
(ii) During the downward motion of the ball after reaching the highest point position, velocity and acceleration all the three quantities are positive.
(d) Initial speed of the ball, u= -29.4 m/s

Final velocity of the ball, v = 0

Acceleration = 9.8 m/s2

Applying in the equation v2 – u2 = 2gs

0 – (-29.4)2 = 2 (9.8) s

s = – 864.36/19.6 = – 44.1

Height to which the ball rise = – 44.1 m (negative sign represents upward direction)

Considering the equation of motion

v = u + at

0 = (-29.4) + 9.8t

t = 29.4/9.8 = 3 seconds

Therefore, the total time taken for the ball to return to the player’s hands is 3 +3 = 6s

Q11. 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 (if the particle rebounds instantly with the same speed, it implies infinite acceleration which is unphysical)

(d) False (true only when the chosen positive direction is along the direction of motion)

Q.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.

Ans:

Height from which the ball is dropped = 90 m

The initial velocity of the ball, u = 0

Let v be the final velocity of the ball

Using the equation

v2 – u2 = 2as ——–(1)

v12 – 0 = 2 x 10 x 90

v1= 42.43 m/s

Time taken for first collision can be given by the equation

v = u + at

42.43 = 0 + (10) t

t1 = 4.24 s

The ball losses one-tenth of the velocity at collision. So, the rebound velocity of the ball is

v2= v – (1/10)v

v2 = (9/10) v

v2= (9/10) (42.43)

= 38.19 m/s

Time taken to reach maximum height after the first collision is

v = u + at

38.19 = 0 + (10)t2

t2 = 3.819 s

Total time taken by the ball to reach the maximum height is

T = t1 + t2

T = 4.24 + 3.819  = 8.05 s

Now the ball will travel back to the ground in the same time as it took to reach the maximum height = 3.819 s

Total time taken will be, T = 4.24 + 3.819 + 3.819 = 11.86

Velocity after the second collision

v3 = (9/10) (38.19)

v3 = 34.37 m/s

Using the above information speed- time graph can be plotted Q13. Provide clear explanations and examples to distinguish between:

( a ) The total length of a path covered by a particle and the magnitude of displacement over the same interval of time.

( b ) The 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].

In ( a ) and ( b ) compare and find which among the two quantity is greater.

When can the given quantities be equal? [For simplicity, consider one-dimensional motion only].

Ans.

( a ) Let us consider an example of a football, it is passed to player B by player A and then instantly kicked back to player A along the same path. Now, the magnitude of displacement of the ball is 0 because it has returned to its initial position. However, the total length of the path covered by the ball = AB +BA = 2AB. Hence, it is clear that the first quantity is greater than the second.

( b ) Taking the above example, let us assume that football takes t seconds to cover the total distance. Then,

The magnitude of the average velocity of the ball over time interval t = Magnitude of displacement/time interval

= 0 / t = 0.

The average speed of the ball over the same interval = total length of the path/time interval

= 2AB/t

Thus, the second quantity is greater than the first.

The above quantities are equal if the ball moves only in one direction from one player to another (considering one-dimensional motion).

Q.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. 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 the magnitude of average velocity. You would not like to tell the tired man on
his return home that his average speed was zero !]

Ans:

Distance to the market = 2.5 km = 2500 m

Speed of the man  walking to the market= 5 km/h = 5 x (5/18) = 1.388 m/s

Speed of the man walking when he returns back home = 7.5 km/h = 7.5  x (5/18) = 2.08 m/s

(a) Magnitude of the average speed is zero since the displacement is zero

(b)

(i) Time taken to reach the market = Distance/Speed = 2500/1.388 = 1800 seconds = 30 minutes

So, the average speed over 0 to 30 minutes is 5 km/h or  1.388 m/s

(ii) Time taken to reach back home = Distance/Speed = 2500/2.08 = 1200 seconds = 20 minutes

So, the average speed is

Average Speed over a interval of 50 minutes= distance covered/time taken = (2500 + 2500)/3000 = 5000/3000 = 5/3 = 1.66 m/s

= 6 km/h

(ii) Average speed over an interval of 0 – 40 minutes = distance covered/ time taken = (2500+ 1250)/2400 = 1.5625 seconds = 5.6 km/h

Q. 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 the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?

Ans:

Instantaneous velocity and instantaneous speed are equal for a small interval of time because the magnitude of the displacement is effectively equal to the distance travelled by the particle.

Q.16. Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent the one-dimensional motion of a particle. Ans:

None of the four graphs shows a one-dimensional motion.

(a) Shows two positions at the same time, which is not possible.

(b) A particle cannot have velocity in two directions at the same time

(c) Graph shows negative speed, which is impossible. Speed is always positive

(d) Path length decreases in the graph, this is also not possible

Q.17. The figure shows the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph shows 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 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 does not represent the path of the particle.

A suitable physical context for the graph can be the particle is dropped from the top of a tower at t =0.

Q. 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 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:

Speed of the police van = 30 km/h = 30 x (5/18) = 25/3 m/s

Speed of a thief’s car = 192 km/h = 192 x (5/18) = 160/3 m/s

Muzzle Speed of  the bullet = 150 m/s

Speed of the bullet = speed of the police van + muzzle speed of  the bullet

= (25/3)+ 150 = 475/3 m/s

The relative velocity of the bullet w.r.t the thief’s car is

v = Speed of the bullet – Speed of a thief’s car

= (475/3) – (160/3) = 105 m/s

The bullet hits the thief’s car at a speed of 105 m/s

Q. 19. Suggest a suitable physical situation for each of the following graphs Ans:

(a)The graph is similar to kicking a ball and then it hits the wall and rebounds with a reduced speed. The ball then moves in the opposite direction and hits the opposite wall that stops the ball.
(b) The graph is showing a continuous change in the velocity of the object and at some instant it losses some velocity. Therefore, it may represent a situation where a ball falls on the ground from a certain height and rebounds with a reduced speed.
(c) A cricket ball moving with a uniform speed hit by a bat for a very short time interval.

Q. 20. The following 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. Ans.

In S.H.M., acceleration, a = – ω2 x , ω is the angular frequency —-(1)
(i) At t = 0.3 s, x < 0 i.e., Position is negative. Moreover, as x is becoming more negative with time, it shows that velocity is negative (i.e., v < 0). However, using equation (1), acceleration will be positive.
(ii) At t = 1.2 s,  Positions and velocity will be positive. Acceleration will be negative.
(iii) At t = -1.2 s, Position, x is negative.  Velocity and acceleration will be positive.

Q.21. The 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. Ans.

Interval 3 is the greatest and 2 is the least. The average velocity is positive for interval 1 and 2 and it is negative for interval 3.

Q. 22. The following 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 points A, B, C and D? Ans.

The change in the speed with time is maximum in interval 2. Therefore, the average acceleration is greatest in magnitude in interval 2.

The average speed is maximum in interval 3

The sign of velocity is positive in interval 1, 2 and 3. The acceleration depends on the slope. The acceleration is positive in interval 1 and interval 3 as the slope is positive. The acceleration is negative in interval 2 as the slope is negative.

Acceleration at A, B, C and D is zero since the slope is parallel to the time axis at these instants.

Q23. 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 nth 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.

For a straight line, the distance covered by a body in nth second is :
SN = u + a (2n – 1)/2     . . . . . . . . ( 1 )
Where,

a = Acceleration

u = Initial velocity
n = Time = 1, 2, 3, . . . . . , n
In the above  case,
a = 1 m/s2 and u = 0.
∴ SN = (2n – 1) / 2     . . . . . . . . . . .( 2 )
This relation shows that:
SN ∝ n                             . . . . .. . . . . ( 3 )

Now substituting different values of n in equation ( 2 ) we get:

 n 1 2 3 4 5 6 7 8 9 SN 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 This plot is expected to be a straight line.

Q.24. 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-1. 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-1 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:

Initial velocity of the ball, u = 49 m/s

Case : I

The boy throws the ball upwards when the lift is stationary. The vertically upward direction is taken as the positive direction. The displacement of the ball is zero.

Considering the equation of motion

s=ut + (1/2)at2

0 = (49)t + (1/2) (-9.8)t2

t = (49 x 2)/9.8 = 98/9.8 = 10 sec

Case : II

As the lift starts moving with a speed of 5 m/s, the initial speed of the ball will be 49 m/s + 5 m/s = 54 m/s

The displacement of the ball will be s =  5t’

Therefore, the time taken can be calculated using the formula

s = ut + (1/2) at2

5t’ = (54) t’ + (1/2)(-9.8) t’2

t’ = 2(54 – 5)/9.8 = 10 sec

The time taken will remain the same in both the cases.

Q. 25. On a long horizontally moving belt figure, 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? Ans:

Speed of child = 9 km h-1

Speed of belt = 4 km h-1
(a) When the boy runs in the direction of motion of the belt, then his speed as observed by the stationary observer = (9 + 4) km h-1 = 13 km h-1.

(b) When the boy runs opposite to the direction of motion of the belt, then speed of child as observed by the stationary observer = (9 – 4) km h-1 = 5 km h-1

(c) Distance between the two parents = 50 m = 0.05 km

Speed of the boy as observed by both the parents is 9 km h-1

Time taken by the boy to move towards one of the parents =0.05 km/9k h-1=0.0056 h =20 S

Q26. Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1 and 30 m s–1. 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 s–2. Give the equations for the linear and curved parts of the plot. Ans.

For the first stone:

Given,

Acceleration, a = –g = – 10 m/s2
Initial velocity, uI = 15 m/s

Now, we know
s1 = s0 + u1t + (1/2)at2
Given, height of the tree, s0 = 200 m
s1 = 200 + 15t – 5t2      . . . . . . . . .  . ( 1 )
When this stone hits the jungle floor, s1 = 0
∴– 5t+ 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 thrown at time t = 0, the negative sign is not possible
∴t = 8 s
For second stone:

Given,

Acceleration, a = – g = – 10 m/s2
Initial velocity, uII = 30 m/s

We know,
s2 = s0 + uIIt + (1/2)at2
= 200 + 30t – 5t2 . . . . . . . . .  . . .  . ( 2 )
when this stone hits the jungle floor; s2 = 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 not possible
∴ t = 10 s
Subtracting equations ( 1 ) from equation ( 2 ), we get
s2 – s1 = (200 + 30t -5t2) – (200 + 15t -5t2)
s2 – s1 =15t                                           . . .  . . . . . . . . . .. . . . . . ( 3 )
Equation ( 3 ) represents the linear trajectory of the two stone, because to this linear relation between (s– s1) and t,,  the projection is a straight line till 8 s.
The maximum distance between the two stones is at t = 8 s.
(s2 – s1)max = 15× 8 = 120 m
This value has been depicted correctly in the above graph.
After 8 s, only the second stone is in motion whose variation with time is given by the quadratic equation:
s2 – s= 200 + 30t – 5t2
Therefore, the equation of linear and curved path is given by :
s– s1 = 15t (Linear path)

s2 ­– s1 = 200 + 30t – 5t2 (Curved path)

Q. 27. The speed-time graph of a particle moving along a fixed direction is shown in the figure. Obtain the distance traversed by the particle between

(a) t = 0 s to 10 s,
(b) t = 2 s to 6 s. Ans:

(a) Distance traversed by the particle between t = 0 s and t = 10 s

= area of the triangle = (1/2) x base x height

= (1/2) x 10 x x12 = 60 m

Average speed of the particle is 60 m/ 10 s = 6 m/s

(b) The distance travelled by the particle between t = 2 s and t = 6 s

= Let S1 be the distance travelled by the particle in time 2 to 5 s and S2 be the distance travelled by the particle in time 5 to 6 s.

For the motion from 0 sec to 5 sec

Now, u = 0 , t = 5 , v = 12 m/s

From the equation v = u + at we get

a = (v – u)/t = 12/ 5 = 2.4 m/s2

Distance covered from 2 to 5 s, S1 = distance covered in 5 sec – distance covered in 2 sec

= (1/2) a (5)2 – (1/2) a (2)2 = (1/2) x 2.4 x (25 – 4) = 1.2 x 21 = 25.2 m

For the motion from 5 sec to 10 sec , u = 12 m/s and a = -2.4 m/s2

and t = 5 sec to t = 6 sec means n = 1 for this motion

Distance covered in the 6 the sec is S2 = u + (1/2) a (2n – 1)

= 12 – (2.4/2) (2 x 1 – 1) = 10.8 m

Therefore, the total distance covered from t = 2 s to 6 s = S1 + S2

= 25.2 + 10.8 = 36 m

Q.28. The velocity-time graph of a particle in one-dimensional motion is shown in the figure.

(a) x (t2) = x (t1) + v (t1) (t2 – t1) + (1/2) a(t2 – t1)2

(b) v(t2) = v(t1) + a(t2 – t1)

(c) Vaverage =  [ x(t2) – x (t1)] /(t2 – t1)

(d)  aaverage =  [ v(t2) – v (t1)] /(t2 – t1)

(e) x (t2) = x (t1) + vav (t2 – t1) + (1/2) aav (t2 – t1)2

(f) x(t2) – x (t1) = Area under the v-t curve bounded by t- axis and the dotted lines. Ans:

The graph has a non-uniform slope between the intervals t1 and t2 (since the graph is not a straight line). The equations (a), (b) and (e) does not describe the motion of the particle. Only the relations (c), (d) and (f) are correct.

 Also Access NCERT Exemplar for Class 11 Physics Chapter 3 CBSE Notes for Class 11 Physics Chapter 3

For students of Class 11, the topics that come under the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line are important as they lay out the foundation for the topics to follow in the future. In order to understand advanced concepts of physics, getting a grasp on the fundamental topics like the one in this chapter is very crucial.

This chapter will be covering simple but essential topics such as comparing the objects as a point object. We will be plotting graphs and determine values according to it. When you walk at a certain speed and then climb down a bus at a certain distance then what will be the (x-t) graph of your motion, check out below how to plot the graph.

Have you ever seen a drunk guy walking some steps forward and a few steps backwards, have you wondered what will be the (x-t) motion graph of his motion? We will be finding the speed of a supercar which will go in front of you, the time taken by a biker to stop when he is stopped by a policeman, know the answer to these questions here. Are you afraid of snakes? Want to know who will catch its prey first among two anacondas racing for their prey, find it here.

Do you love basketball? Want to know at what direction the ball is accelerating when it is thrown upwards by the referee and also find out here what will be the ball’s acceleration and velocity at the highest point. Let’s see what statements are true regarding the one-dimensional motion of an object. Have you seen a rubber ball bounce back when you throw it down the ground, what will be the speed-time graph of that, see the answers below. You can check out NCERT Solutions for Class 11 Physics for more chapter-wise solutions.

We will be providing the explanation for the difference between the total length of the path covered by a particle and magnitude of displacement in the same interval of time and we will also find out the difference between average velocity and average speed in the same time interval. Do you know there is no difference between instantaneous speed and velocity, find out why here? This chapter has a vast number of questions on every important topic in this chapter.

## Subtopics of Class 11 Physics Chapter 3 Motion in a Straight Line

1. Introduction
2. Position, path length and displacement
3. Average velocity and average speed
4. Instantaneous velocity and speed
5. Acceleration
6. Kinematic equations for uniformly accelerated motion
7. Relative velocity

Here we have provided students with NCERT Solutions for Class 11 Physics Chapter 3 pdf to help them learn more effectively and understand the basic concepts of physics. Solving these questions would ensure better clarity.

NCERT solutions for Class 11 physics Chapter 3 pdf can be accessed anytime and can be downloaded easily.  CoolGyan’S presents the best study materials, notes, study material, books, previous year question papers, sample papers, videos and animation to help the students in their all-important Class 11 and entrance examinations.

## Frequently Asked Questions on NCERT Solutions for Class 11 Physics Chapter 3

### Why should I use the NCERT Solutions for Class 11 Physics Chapter 3 while preparing for the exams?

The NCERT Solutions for Class 11 Physics Chapter 3 provides students with simple and step wise answers to the questions present in the textbook. The solutions are helpful to understand the method of answering the questions in the board exams. Students can save their time and fasten their revision which is very effective for exam preparation. The expert teachers at CoolGyan’S have designed the solutions based on the CBSE guidelines and syllabus.

### What are the topics covered under the Chapter 3 of NCERT Solutions for Class 11 Physics?

The topics covered under the Chapter 3 of NCERT Solutions for Class 11 Physics are –
Introduction
Position, path length and displacement
Average velocity and average speed
Instantaneous velocity and speed
Acceleration
Kinematic equations for uniformly accelerated motion
Relative velocity.

### What is the meaning of acceleration in Chapter 3 of NCERT Solutions for Class 11 Physics?

Acceleration is defined as the rate of change of velocity with respect to time. Acceleration is a vector quantity as it has both magnitude and direction. It is also the second derivative of position with respect to time or it is the first derivative of velocity with respect to time..

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