Case Study Questions Class 11 Physics Motion in a Straight Line

Case Study Questions Class 11 Physics Chapter 3 Motion in a Straight Line

CBSE Class 11 Case Study Questions Physics Motion in a Straight Line. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Motion in a Straight Line .

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Physics Motion in a Straight Line

 

Case Study – 1

If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of actual path length travelled by object. It is scalar quantity having SI unit of metre while displacement refers to the shortest distance between initial and final position of object. It is vector quantity. The magnitude of the displacement for a course of motion may be zero but the corresponding path length is not zero. using this data answer following questions.

 

1) Can path length be zero for motion of body from one point to other point?

a) Yes

b) No

2) For any given motion from point A to B, displacement =10m and distance = 5m. Is it possible?

a) Yes

b) No

3) For rectilinear motion displacement can be

a) Positive only

b) Negative only

c) Can be zero

d) All of the above

4) Define distance and displacement of particle.

5) Write difference between distance and displacement.

 

Answer key – 1

1) b

2) b

3) d

4) Distance and displacement are two quantities that seem to mean the same but are different with different meanings and definitions. Distance is the measure of “how much distance an object has covered during its motion” while displacement refers to the measure of “how far the abject actually from initial place.”

5) difference between distance and displacement is given by

No.

Distance

Displacement

1 The complete length of the path between any two points is called distance. Displacement is the shortest length between any two points.
2 Distance is a scalar quantity Displacement is a vector quantity
3 For any given motion distance is always greater than or equal to displacement For any given motion displacement is always smaller than or equal to distance.
4 The distance can only have positive values. Displacement can be positive, negative, and even zero.

 

Case Study – 2 

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below:

v = u + at           s = ut + ½ at2           2a s = v2 – u2

Where u is the initial velocity of the object which moves with uniform acceleration a for Time t, v is the final velocity and s is the distance travelled by the object in time t.

 

1) equation of motions are applicable to motion with

a) uniform acceleration

b) non uniform acceleration

c) constant velocity

d) none of these

2) There are 4 equation of motion. True or false?

a) True

b) False

3) The brakes applied to a car produce an acceleration of 10 m/s2 in the opposite direction to the motion. If the car takes 1 s to stop after the application of brakes, calculate the distance traveled during this time by car.

4) An object is dropped from a tower falls with a constant acceleration of 10 m/s2. Find its speed 10 s after it was dropped.

5) A bullet hits a Sand box with a velocity of 10 m/s and penetrates it up to a distance of 5 cm. Find the deceleration of the bullet in the sand box

 

Answer key – 2

1) a

2) b

3) Here in this problem,

v = 0

a = -10 m/s2 (as acceleration is retarding)

t = 1 sec.

To find: distance travelled

Solution: using kinematic equation

v = u + at

0=  u + -10*1

u = 10 m/s

Therefore distance is given by

s = ut + ½ at2

s = 10*1-(1/2)*10*12

s= 5m

 

4) here in this problem,

u= 0

a= 10 m/s2(as acceleration is in the direction of gravity)

t= 10 sec.

To find: final velocity after 10 second

Solution: using kinematic equation

v = u + at

v = 0+ 10*10

v= 100 m/s

 

5) Here in this problem,

v= 0(as bullet is going to stop)

u= 10 m/s

s= 5m.

To find: deceleration of the bullet

Solution: using kinematic equation

2a s = v2 – u2

2*a*5= 02-102

10a=-100

A=-100/10

a = -10m/s2. Negative sign indicates that it is deceleration.

 

Case Study – 3

The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval Dt becomes infinitesimally small. In other words

The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the y-axis. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement and slope of velocity time graph represents acceleration of object.

The variation in distance with time for an object moving in a straight line can be represented by a position-time graph. In this graph, time is represented along the x-axis and the displacement is represented along the y-axis. Answer the following questions based on paragraph given.

1) The area under velocity time graph gives

a) Displacement over given time interval

b) Acceleration

c) Velocity

d) None of these

2) Slope of velocity time graph gives

a) Acceleration

b) Velocity

c) Distance

d) Displacement.

3) Write note on velocity time graph.

4) Write a note on position time graph

5) What is instantaneous velocity ?

 

Answer key-3

1) a

2) a

3) The change in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis and the velocity is represented along the y-axis. slope of velocity time graph represents acceleration of object The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement and

4) The change in distance with time for an object moving in a straight line can be given by a position-time graph. In this graph, time is represented along the x-axis and the displacement is represented along the y-axis.

5) The velocity at an instant is defined as the limit of the average velocity as the time interval t becomes infinitesimally small. In other words

Case Study – 4

Introduce the concept of relative velocity.

Relative velocity is velocity of any object with respect to other object which may be stationary or moving. Consider two objects A and B moving uniformly with average velocities vA and vB in one dimension, say along x-axis. (Unless otherwise specified, the velocities mentioned in this chapter are measured with reference to the ground). If xA (0) and xB (0) are positions of objects A and B, respectively at time t = 0, their positions xA (t) and xB (t) at time t are given by

xA (t) = xA (0) + vA t

xB (t) = xB (0) + vB t

Then, the displacement from object A to object B is given by

xBA(t) = xB (t) – xA (t)

= [xB (0) – xA (0) ] + (vB – vA) t.

It tells us that as seen from object A, object B has a velocity vB – vA because the displacement from A to B changes steadily by the amount vB – vA in each unit of time. We say that the velocity of object B relative to object A is vB – vA

VBA = vB – vA

Similarly, velocity of object A relative to object B is:

VAB = vA – vB

This shows VBA= – VAB.

1) Velocity of object A relative to object B is

a) VAB = vA – vB

b) VBA = vB – vA

c) None of these

2) Velocity of object B relative to object A is

A) vB – vA

b) vA – vB

c) None of these

3) What is relative velocity?

4) What is relative displacement?

5) Show that VBA= – VAB

 

Answer key-4

1) a

2) a

3) Relative velocity is velocity of any object with respect to other object which may be stationary or moving.

4) If xA (0) and xB (0) are positions of objects A and B, respectively at time t = 0, their positions xA (t) and xB (t) at time t are given by

xA (t) = xA (0) + vA t

xB (t) = xB (0) + vB t

Then, the displacement from object A to object B is given by

xBA(t) = xB (t) – xA (t).

 

5) By definition of relative velocity We say that the velocity of object B relative to object A is vB – vA

VBA = vB – vA

Similarly, velocity of object A relative to object B is:

VAB = vA – vB

This shows VBA= – VAB.

 

Case Study – 5

When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity average velocity. Average velocity is defined as the change in position or displacement (x) divided by the time intervals (t), in which the displacement occurs:

Where x2 and x1 are the positions of the object at time t2and t1, respectively. The SI unit for velocity is m/s or m s–1, although km h–1 is used in many everyday applications. Like displacement, average velocity is also a vector quantity. Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place:

Average speed = Total path length/ Total time interval.

Average speed has obviously the same unit (m s–1) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length.

The velocity at an instant is defined as the limit of the average velocity as the time interval Dt becomes infinitesimally small. In other words

Note that for uniform motion, velocity is the same as the average velocity at all instants. Instantaneous acceleration is defined in the same way as the instantaneous velocity

1) For uniform motion instantaneous velocity is same as

a) Average velocity

b) Average acceleration

c) Instantaneous speed

d) None of these

2 If velocity is constant then

a) Acceleration is zero

Acceleration is positive

c) Acceleration is negative

d) None of these

3) Define average speed

4) Define instantaneous acceleration

5) Define average velocity

 

Answer key-5

1) a

2) a

3) Average speed is defined as the total path length travelled divided by the total time.

Average speed= Total path length / Total time interval.

Average speed has SI unit of m/s. it is scalar quantity it has only magnitude and doesn’t have any direction. it is always positive.

4) Instantaneous acceleration is defined rate of change of velocity with time when time tends to zero

5) Average velocity is defined as the change in position or displacement (Dx) divided by the time intervals (Dt), in which the displacement occurs :

Where x2 and x1 are the positions of the object at time t2 and t1 , respectively. The SI unit for velocity is m/s or m s–1.

Updated: March 22, 2022 — 2:46 pm

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