Motion

Displacement refers to the measure of the shortest path between any two points whereas distance is the measurement of path between two points. The SI unit of both distance and displacement is meter (m).

Displacement is a vector quantity as it has both magnitude and direction.

Displacement along a circular path is zero as the final and the initial points meet each other that is both the points are same.

(iii) An athlete completes one lap in a race

As the track is circular, the displacement would be zero but distance will be equal to the circumference of the track.

(i) Distance would be PQ+QR = 12m whereas displacement is path PR= 8m as displacement is the shortest distance between two points.

(ii) In this case displacement is zero as initial and final point is same but distance would be PQ+QP= 20m.

Average speed of the object in an interval of time is the distance travelled by the object divided by the duration of interval of time. Average speed is a scalar quantity while average velocity is a vector quantity.

Maximum speed = 80 km/hr

Average speed =

= = 40 km/hr

Ratio = = = 2

Average speed is a scalar quantity and hence it cannot be zero but average velocity is a vector quantity and it can be zero if the object is moving in a circular path.

(i) Acceleration depends on the change in the velocity and the time taken by the object to travel.

(ii) Acceleration =

=

SI unit of acceleration is meter per second square (m/s²).

(i) Retardation is the term used for negative acceleration.

Explanation: If the velocity of an object is decreasing, it is said to have negative acceleration.

(ii) A bus starting from rest attains a velocity of 54 km/h in 60 s, its acceleration is 0.25 m/s².

Explanation: initial velocity = 0

time = 60 secs

final velocity = 54 km/hr = m/s = 15 m/s

Acceleration =

=

= = 0.25 m/s²

The velocity of the particle will be zero because the distance is constant and the time is increasing and this will happen only when the particle is stationary.

Average speed =

= + =10 m/s + 10 m/s = 20 m/s

Similarity is that both the graphs shows uniform acceleration and dissimilarity is that in the first graph the object starts from rest whereas in the second graph the object has some initial velocity.

The acceleration would be the area under the graph.

Acceleration = = = -1.25 m/s²

Graph P is having negative acceleration as the time is increasing, velocity is decreasing and finally comes to rest which shows retardation whereas Graph Q represents positive acceleration as the time is increasing velocity is also increasing.

(i) The object in Graph Q starts from rest as when time is zero, velocity is also zero.

(ii) Graph P will have more velocity after 5 secs as from the graphs we can infer that in Graph P after 5 secs the velocity is 14 m/s whereas in Graph Q the velocity reaches 12 m/s. Both the graph shows uniform acceleration but object in graph P has an initial velocity of 2 m/s and object in graph Q starts from rest.

Using the first equation of motion: v = u + at

where: v = final velocity
u = initial velocity = 10 m/s
a = acceleration = 2.5 m/s²
t = time = 10s

v = 10 m/s + 2.5 m/s² x 10s = 10+25 m/s = 35 m/s

Using the second equation of motion: s = ut + at²

where: s = distance covered =50 m
u = initial velocity = ?
a = acceleration = 5 m/s²
t = time = 4s

s = ut + at²

50 = u x 4s + x 5 x 4²

50 = u x 4s + 40

50 – 40 = u x 4s

10 = u x 4s

u = m/s = 2.5 m/s

Using the third equation of motion: v² – u² = 2as

where: s = distance covered =?
v = final velocity = 18 m/s
u = initial velocity = 14 m/s
a = acceleration = 6 m/s²
18² – 14² = 2×6× s

324 – 196 = 12 × s
= s

s = 10.67 m

We have to first find the retardation.
Using the third equation of motion: v² – u² = 2as
where: s = distance covered = 10 m
v = final velocity = 0 m/s (brakes are applied so that bus
comes to rest)
u = initial velocity = 72 km/h = 20 m/s
a = acceleration = ? m/s²
v² – u² = 2as

0 – 20² = 2 × a × 10

-400 = 20 × a

= a

a = -20 m/s²

Now, u = initial velocity = 144 km/h = 40 m/s
s = distance covered = ?? m
v = final velocity = 0 m/s ( brakes are applied so that bus comes to rest)
a = acceleration = -20 m/s²

Using the third equation of motion: v² – u² = 2as

0 – 40² = 2 × -20 × s

-1600 = -40 × s

= s

s = 40 m

Uniform circular motion is defined as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. In all instances, the object is moving tangentially to the circle. For example an artificial satellite moving along its axis around the earth, the car taking a curve in a circular track.

The particle is moving in a circular path because at every point the direction is changing along the path and is always tangential to the circular path.

Uniform Motion means that the object covers the equal distance in equal interval of time that is the speed is constant. So path can be straight or curved but the magnitude remains the same.

The given quantity represents change of velocity per unit time that is acceleration and its magnitude is 10 m/s².

The area under the velocity time graph gives the displacement of the object and its value is 40 m.

The motion of second’s hand represent uniform speed because as it is moving in a circular motion, its direction is changing at every point and hence uniform velocity will not be the case.

Using Pythagoras theorem: PQ² + QR² = PR²

4² + 3² = PR²
PR² = 16 + 9 = 25

PR = 5 km

A body moving in a circular path having uniform speed is a true example where speed remains constant but velocity changes because every point, the direction changes at every point and is tangentially to the circular path.

(i)

(iii) Acceleration of car = slope of line AB = = = = 0.5 m/s²

(i) Speed is equal to the slope of the distance-time graph.

Slope = Speed=

(ii) Acceleration is also the slope of the velocity-time graph.

Slope = Acceleration=

(iii) The Magnitude of Displacement is the area under the curve of velocity time graph.

Velocity is equal to the slope of the distance-time graph.

Slope = Velocity=

= m/s

= m/s =4 m/s

The graph represents a straight line which means that the object is moving with uniform acceleration.

Acceleration= = = m/s²

(i) If the line is parallel to X-axis in a velocity-time graph that means that with increase in time velocity remains constant. There will be no acceleration and the body will move with constant velocity. It cannot be parallel to Y-axis because it will mean that time is constant but velocity is increasing which is practically not possible.

(ii)

The first graph represents a body which is moving with a uniform acceleration and the second graph represents a body which is moving with uniform retardation.

(iii) Initial velocity = 36 km/h = 10 m/s

Final velocity = 54 km/h = 15 m/s

Acceleration= = = = 0.5 m/s²

(i) Acceleration from A to B = = = 8.33 m/s²
(ii) Acceleration from B to C = = -10 m/s²

(iii) Distance covered in the region ABE = × base × height

= × 3s × 25 m/s

= 37.5 m

(iv) Average velocity from C to D =

= = 7.5 m/s

(v) Distance covered in the region BCFE = × sum of parallel sides x distance between them = × ( 25+15) × 1 = 20 m

(i) moving with a uniform retardation

(ii) moving with a variable acceleration

Using the first equation of motion: v = u + at

where: v = final velocity = 0 m/s
u = initial velocity = 0.5 m/s
a = acceleration = -0.105 m/s²
t = time = ??

v = u + at
0 = 0.5 - 0.105 x t
0.105 x t = 0.5
t = s =4.76 s

(i) The stone is moving with circular motion.

(ii) Yes it is an example of accelerated motion. Though the velocity is constant but as the thread is moving it is changing its direction at each point which is tangential to the circular path.

(iii) Centripetal force keeps the stone in its path.

(iv) The force is directed towards the centre of the circular path.

a) From the graph we can infer that in the interval OA, it’s almost a straight line with a positive slope. Therefore velocity is positive that is with increase in time distance also increases and thus acceleration is also positive as well as uniform.

b) In the interval AB, the line has a negative slope as with increase in time distance decreases. Here velocity is negative and it shows retardation.

c) BC represents more downfalls in the slope of the line which means velocity will become more negative and hence it shows retardation with greater magnitude than AB.

d) CD shows that with increase in time distance is increasing again with a positive slope. Therefore velocity is positive which implies acceleration is also positive.

a) The first equation of motion: v = u + at

where: v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)
t = time (s)

b) The second equation of motion: s = ut + at²

where: s = distance covered (m)
u = initial velocity (m/s)
a = acceleration (m/s²)
t = time (s)

c) The third equation of motion: v² – u² = 2as

where: s = distance covered (m)
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration (m/s²)

Using the second equation of motion: s = ut + at²

where: s = distance covered = 20 m
u = initial velocity = 0 m/s
a = acceleration = ? m/s²
t = time = 2s

s = ut + at²
20 = 0 × 2 + × a × 2²
20 = 0 + 2 × a
a = 10 m/s²

Final velocity, v after 2s

The first equation of motion: v = u + at

where: v = final velocity = ?
u = initial velocity = 0 m/s
a = acceleration = 10 m/s²
t = time = 2s

v = u + at
v = 0 + 10 × 2 = 20 m/s

Now it is given that in next 4s it covers 160 m. But now the vehicle has gained some velocity. So the final velocity of the previous case will become the initial velocity in this case.

Using the second equation of motion: s = ut + at²

where: s = distance covered = 160 m
u = initial velocity = 20 m/s
a = acceleration = ? m/s²
t = time = 4s

s = ut + at²
160 = 20 × 4 + × a × 4× 4
160 = 80 + a × 8
160 – 80 = a × 8
80 = a × 8
a = 10 m/s²

This shows that acceleration is uniform.

The first equation of motion: v = u + at

where: v = final velocity = ?
u = initial velocity = 0 m/s
a = acceleration = 10 m/s²
t = time = 7s

v = u + at
v = 0 + 10 × 7
v = 70 m/s

Using the second equation of motion: s = ut + at²

where: s = distance covered
u = initial velocity
a = acceleration
t = time = 4s

Distance covered in 4s :

s₁ = ut + at²
s₁ = u × 4 + a × 4²
s₁ = 4u + a ………………… equation 1

Distance covered in 5s:

s₂ = ut + at²
s₂ = u × 5 + a × 5²
s₂ = 5u + a …………………… equation 2

Distance travelled by an object between 4^th and 5^th sec:

equation 2 – equation 1
(s₂ – s₁ ) = (5u + a ) - (4u + a)
(s₂ – s₁ ) = 5u – 4u +a - a)
(s₂ – s₁ ) = u +a

(i) Over speeding on roads increases the chances of accidents which could be fatal also. Some serious injuries may happen due to this. Also they could hit some other car on the road as well.

(ii) Sharman behaved as a responsible citizen and patiently he convinced his friend not to drive like this. He is smart, intelligent, responsible as well as patient.

(iii) Speed = 180 km/h

1 km = 1000 m and 1 hour = 3600 secs
speed = = = 5 m/s

(iv) The first equation of motion: v = u + at

where: v = final velocity = 5 m/s
u = initial velocity = 0 m/s
a = acceleration = ? m/s²
t = time = 20s

v = u + at

5 = 0 + a × 20

a = = 0.25 m/s²

(v) I would also suggest him to drive in the speed limit and try to make him realize the importance of life.

(i) Rajnish gave the correct explanation as for a uniform motion the velocity should be uniform. But the metro has stopped at various stations which means that it had come to rest and its velocity at that point is zero.

(ii) Manoj had used metro which is free from the traffic jam and thus manages to reach early.

(iii) Manoj is a responsible citizen as he had used a public transport instead of the personal car which helps in conservation of fuel.

(i) Neha is not concerned about the traffic rules and hence does not care about the problems faced by other people. On the other hand, Priya being a responsible citizen follows the traffic rules. She understands that car being parked in a tow away zone can lead to traffic jams and can lead to chaos.

(ii) Non-uniform motion is exhibited by the car in the market area as Neha has to sometimes stop the car, sometimes move the car. The velocity is not uniform.

(iii) With the increase in time, distance remains the same which implies that the object is at rest.

For the first 40 m:

Using the third equation of motion: v² – u² = 2as

where: s = distance covered = 40 m
v = final velocity = ? m/s
u = initial velocity = 0 m/s
a = acceleration = 1 m/s²

v² – u² = 2as
v² – 0 = 2 × 1 × 40
v² = 80
v = 8.94 m/s

Using the first equation of motion: v = u + at

where: v = final velocity = 8.94 m/s
u = initial velocity = 0 m/s
a = acceleration = 1 m/s²
t₁ = time = ?s

v = u + at₁
8.94 = 0 + 1 × t₁
t₁ = 8.94s

Here its given the speed is uniform not velocity, therefore
acceleration will be 1 m/s² and speed will be 8.94 m/s everywhere.

Using the third equation of motion: v² – u² = 2as
where: s = distance covered = 60 m
v = final velocity = ?? m/s
u = initial velocity = 8.94 m/s
a = acceleration = 1 m/s²
v² – u² = 2as
v² – (8.94)² = 2 × 1 × 60

v²- 80 = 120
v² = 120 + 80 = 200
v = 14.14 m/s

Using the first equation of motion: v = u + at

where: v = final velocity = 14.14 m/s
u = initial velocity = 8.94 m/s
a = acceleration = 1 m/s²
t₂ = time = ?s

v = u + at₂

14.14 = 8.94 + 1 × t₂

t₂ = 14.14 – 8.94

t₂ = 5.2 secs

Time taken by Q to make one revolution around P = Time taken by Q to complete one revolution – time taken by P to complete one revolution = 5 mins – 2 mins = 3 mins

Helium is less denser than the air and so it is lighter then the air, so when the car is accelerated the air starts moving backwards but as the helium is lighter it is pushed forward by the air collecting at its back.

Average speed = = m/min = 6 m/min

Let the total distance be 100 km

One third of 100 km = 33.33 km

time =

t₁ = = 3.33 hr

Distance left = 100 km – 33.33 km = 66.67 km

One fourth of 66.67 km = 16.67 km

t₂ = = 0.83 hr

Remaining distance = 100 km – 33.33 km – 16.67 km = 50 km

t₃ = = 1.25 hr

t₁+t₂+t₃ = 5.41 hr

Average speed = = = 18.4 km/hr

_{Distance covered by car A in time t is, s1} = 60 t

For car B:

Using the second equation of motion: s = ut + at²

where: s = distance covered = s₂
u = initial velocity = 70 km/h
a = acceleration = -20 km/h
t = time = t

s = ut + at²

s₂ = 70t + (-20) × t²

s₂ = 70t – 10 t²

But it is given that s₂ – s₁ =2.5 km
70t – 10 t² - 60 t = 2.5

10t - 10 t² = 2.5

t – t² = 0.25

t – t² – 0.25 =0

t² –t +0.25 =0

(t – 0.5)² =0

t = 0.5 hr

Using the second equation of motion: s = ut + at²

where: s = distance covered = 75 m
u = initial velocity = 0 m/s
a = acceleration = 1 m/s²
t = time = ts

s = ut + at²
75 = 0 + × 1 × t²

75 × 2 = t²

150 = t²

t = 12.24 s

Using the third equation of motion: v² – u² = 2as
where: s = distance covered = 75 m
v = final velocity = ?? m/s
u = initial velocity = 0 m/s
a = acceleration = 1 m/s²

v² – u² = 2as

v² = 2 × 1 × 75 = 150

v = 12.24 m/s

For second part distance is 25 m.

Using the second equation of motion: s = ut + at²

where: s = distance covered = 25 m
u = initial velocity = 12.24 m/s
a = acceleration = 1 m/s²
t = time = ts

s = ut + at²

25 = 12.24t + 1 x t²

t² + 24.48t – 50 = 0

t = 1.99 s

For bus:

Using the second equation of motion: s = ut + at²

where: s₁ = distance covered = ? m
u = initial velocity = 0 m/s
a = acceleration = 1 m/s²
t = time = ts

s₁ = ut + at²

s₁ = 0 × t + × 1 × t² = t²/2

Distance travelled by the man, s₂ = 10t

s₂ – s₁ = 48 m

10t - t²/2 = 48

20t – t² = 96

20t – t² – 96 = 0

t² – 20t +96 =0

(t-12)(t-8) = 0

t = 12s or 8s

With increase in time distance remains constant which means it is not moving.