Describing Motion Around Us - Chapter 4 Class 9 Exploration

My father went to a shop from home which is located at a distance of 250 m on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?
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He covers home→shop→home→shop→home, i.e. 250 m four times.
Total distance
= 4 × 250 = 1000
m
. He ends where he started, so
displacement = 0
.
Quantity
Value
Total distance travelled
1000 m
Displacement
0 m (returns home)
← Back to: 4.1.2 Distance travelled and displacement

A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is 3 m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.
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Ground floor = 0 m, fourth floor =
4 × 3 = 12
m
, second floor =
2 × 3 = 6
m
.
(i) Total vertical distance = up 12 m + down (12 − 6) m =
12 + 6 = 18
m
.
(ii) Displacement = from ground (0) to second floor (6 m) =
6 m upward
.
Quantity
Value
Total vertical distance
18 m
Displacement
6 m upward
← Back to: 4.1.2 Distance travelled and displacement

A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?
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Yes.
The speedometer shows only the magnitude of velocity (speed). If the scooter
turns
, the
direction
of its velocity changes even though the speed is constant. A change in velocity (here, in direction) means the scooter is accelerating — just like in uniform circular motion.
Quantity
Does it change?
Speed (speedometer)
No — stays constant
Direction of velocity
Yes — on turning
Is it accelerating?
Yes
← Back to: 4.4.1 Uniform circular motion

A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
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u = 0, v = 24
m s
-1
, t = 6
s
.
Acceleration
a =
v - u
t
=
24 - 0
6
= 4
m s
-2
.
Distance
s = ut +
1
2
at
2
= 0 +
1
2
(4)(6
2
) = 72
m
.
Quantity
Value
Average acceleration a
4 m/s²
Distance travelled s
72 m
← Back to: 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
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u = 28
m s
-1
, v = 0, s = 98
m
.
Using
v
2
= u
2
+ 2as
:
0 = 28
2
+ 2a(98) ⇒ a =
-784
196
= -4
m s
-2
.
Using
v = u + at
:
0 = 28 + (-4)t ⇒ t = 7
s
.
Quantity
Value
Acceleration a
−4 m/s²
Time to stop t
7 s
← Back to: 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.
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No.
Both graphs are straight lines, so each object moves with a
constant
velocity equal to the slope of its line. The two slopes are different (the lines are not parallel), so the velocities are never equal. The point where the lines cross only means the two objects are at the
same position
at that instant, not that they have the same velocity.
Aspect
Conclusion
Each graph
A straight line (constant v)
The two slopes
Different
Equal velocity ever?
Never
Where lines cross
Same position only
← Back to: 4.2.2 Position-time graphs

A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).
(i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions.
(ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time.
(iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
(iv) The average speed of A over the 10 s time interval is greater than that of B since B’s speed is lower than A’s in some segments.
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Correct:
(i) and (ii)
.
Both have the same initial and final positions, so their displacement is equal →
average velocity is equal
. Since both move in one direction (positions only increase), the total distance equals the displacement for each, so they also cover
equal distance in equal time
→ average speeds are equal. Hence (iii) and (iv) are wrong.
Statement
Verdict
(i) Average velocities equal
Correct
(ii) Average speeds equal
Correct
(iii) and (iv)
Wrong
← Back to: 4.2.2 Position-time graphs

A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
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u = 54
km h
-1
= 15
m s
-1
, v = 36
km h
-1
= 10
m s
-1
, t = 36
s
.
Distance = average velocity × time
=
15 + 10
2
× 36 = 12.5 × 36 = 450
m
.
Quantity
Value
Initial velocity u
15 m/s
Final velocity v
10 m/s
Time t
36 s
Distance travelled
450 m
← Back to: 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
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Phase 1 (speed up):
s
1
=
1
2
(0+20)(5) = 50
m
.
Phase 2 (constant):
s
2
= 20 × 10 = 200
m
.
Phase 3 (brake):
s
3
=
1
2
(20+0)(6) = 60
m
.
Total distance
= 50 + 200 + 60 = 310
m
.
Phase
Distance
Speeding up (5 s)
50 m
Constant speed (10 s)
200 m
Braking (6 s)
60 m
Total
310 m
← Back to: 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 m ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of 2.5 m s⁻². Will the bus be able to stop before reaching the obstacle?
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u = 36
km h
-1
= 10
m s
-1
.
Reaction distance (before braking)
= 10 × 0.5 = 5
m
.
Braking distance:
v
2
= u
2
+ 2as ⇒ 0 = 10
2
+ 2(-2.5)s ⇒ s =
100
5
= 20
m
.
Total stopping distance
= 5 + 20 = 25
m
< 30
m
.
Yes, the bus stops in time
, with 5 m to spare.
Distance
Value
Reaction (0.5 s)
5 m
Braking
20 m
Total stopping
25 m
Obstacle ahead
30 m
Stops in time?
Yes — 5 m to spare
← Back to: 4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

A student said, “The Earth moves around the Sun”. In this context, discuss whether an object kept on the Earth can be considered to be at rest.
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Rest and motion are always relative to a chosen
reference point
. With respect to the
Earth’s surface
, the object’s position does not change, so it is
at rest
. With respect to the
Sun
, the Earth (and the object on it) keeps moving, so it is
in motion
. So it can be called at rest or in motion depending on the reference point chosen.
Reference point
The object is
Earth's surface
At rest
The Sun
In motion
← Back to: 4.1.1 Describing position

The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
(i) while cyclist is moving with constant velocity.
(ii) when the velocity of cyclist is decreasing.
Also, calculate the displacement and average acceleration in the 120 s time interval.
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Shading:
the rectangular region under the flat part is the displacement during constant velocity; the region under the down-sloping part is the displacement while velocity is decreasing.
Displacement = area under the whole graph.
Reading Fig. 4.30 (ramp up 0–20 s to 3 m s⁻¹, constant 3 m s⁻¹ for 20–100 s, then falling to about 2 m s⁻¹ by 120 s):
s ≈
1
2
(20)(3) + (80)(3) +
1
2
(3+2)(20) = 30 + 240 + 50 = 320
m
.
Average acceleration
=
v
120
- v
0
120
=
2 - 0
120
≈ 0.017
m s
-2
. (Read the exact values from your copy of Fig. 4.30.)
Quantity
Value
Displacement (area)
≈ 320 m
Average acceleration
≈ 0.017 m/s²
← Back to: 4.2.3 Velocity-time graphs

A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
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Distance =
area under the velocity-time graph
(velocity in km h⁻¹, time in h, so area is in km).
Estimating Fig. 4.31 (rises to about 7.5 km h⁻¹ by 2 h, holds, then eases to about 6.5 km h⁻¹ by 6 h), the average speed is roughly 6.5–7 km h⁻¹ over 6 h, giving a distance of about
40 km
. Count the grid squares under the curve on your copy of Fig. 4.31 for a precise value.
Quantity
Value
Method
Area under the v–t graph
Distance run
≈ 40 km
← Back to: 4.2.3 Velocity-time graphs

On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
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Phase 1 (constant):
s
1
= 6 × 120 = 720
m
.
Phase 2 (accelerating):
s
2
= ut +
1
2
at
2
= 6(6) +
1
2
(1)(6
2
) = 36 + 18 = 54
m
.
Total displacement
= 720 + 54 = 774
m
(the total area under the velocity-time graph).
Phase
Distance
Constant velocity (2 min)
720 m
Accelerating (6 s)
54 m
Total displacement
774 m
← Back to: 4.2.3 Velocity-time graphs

Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned.
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Car A:
a
A
=
5
5
= 1
m s
-2
; displacement in 5 s
=
1
2
(5)(5) = 12.5
m
.
Car B:
a
B
=
3
10
= 0.3
m s
-2
; displacement in 10 s
=
1
2
(3)(10) = 15
m
.
Each displacement is the area of the triangle under that car’s line.
Car
Displacement
Car A (a = 1 m/s², 5 s)
12.5 m
Car B (a = 0.3 m/s², 10 s)
15 m
← Back to: 4.2.3 Velocity-time graphs

Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute’s hand of the wall clock. During the given time interval, what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute’s hand is 7 cm (Fig. 4.32).
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6:00 to 7:30 = 90 min = 1.5 h. The minute hand makes
90
60
= 1.5
revolutions. Radius
R = 7
cm
.
(i) Distance
= 1.5 × 2π R = 1.5 × 2 ×
22
7
× 7 = 1.5 × 44 = 66
cm
.
(ii) After 1.5 turns the tip is diametrically opposite its start, so displacement
= 2R = 14
cm
.
(iii) Speed
=
66
cm
90
min
≈ 0.73
cm min
-1
(i.e. 44 cm h⁻¹).
(iv) Velocity
=
14
cm
90
min
≈ 0.16
cm min
-1
, directed from the start position to the end position.
Quantity
Value
Distance
66 cm
Displacement
14 cm
Speed
≈ 0.73 cm/min
Velocity
≈ 0.16 cm/min
← Back to: 4.4.1 Uniform circular motion

Project 1
Take a cardboard disc (radius ~ 8 cm). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters ‘ABCDEF’ on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different? Justify your answer.
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The outer numbers (7 cm out) move faster than the inner letters (4 cm in), because in one turn the outer marks cover a larger circle (
2π R
is bigger for larger
R
). The faster outer numbers blur and fade sooner as you speed up, while the slower inner letters stay readable longer. So their speeds are
different
even though they complete a turn in the same time.

Project 2
Many smartphones have an inbuilt accelerometer that can detect very small accelerations. Install an app, such as Phyphox (phyphox.org) and open ‘Accelerometer (without g)’. Note the readings when (i) the phone is on an outstretched palm, and (ii) the phone is kept on the floor. What differences do you observe? What does this tell you about motion and acceleration in real situations? (Such tiny, involuntary movements are also studied in medical research, for example, in movement disorders.) This activity is recommended to be performed as a classroom group activity facilitated by a teacher.
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On the palm, the reading flickers around small non-zero values because your hand makes tiny, involuntary movements (accelerations). On the floor it reads close to zero and steady. This shows that real objects are rarely perfectly still — small accelerations are present even when something looks at rest.

Project 3
For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (4.4a) and (4.4b). Using these two equations, three more can be derived, of which we derived one (4.4c). Derive the remaining two equations:
s = vt -
1
2
at
2
and
s =
1
2
(u+v)t
. Using the area of a trapezium, derive the second equation above.
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Trapezium area:
the velocity-time graph is a trapezium with parallel sides
u
and
v
and width
t
:
s =
1
2
(u+v) t
Other equation:
from
v = u + at
we get
u = v - at
. Substituting into
s = ut +
1
2
at
2
:
s = (v-at)t +
1
2
at
2
= vt - at
2
+
1
2
at
2
, so
s = vt -
1
2
at
2

Project 4
Plot graphs for data given in Table 4.4, using different X and Y scales, on different graph papers. Compare the graphs to find how the appearance of the graph is affected by the choice of scales and decide which scale is better and why. Now repeat this with any graph plotting app. Such apps generally automatically adjust the axes to fit the data well on the screen.
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The same data can look steep or flat depending on the scales chosen, though the actual values are unchanged. A good scale spreads the data across most of the page so points are easy to plot and read. Graph-plotting apps auto-fit the axes for exactly this reason.

Project 5
Talk to a motor mechanic about how a vehicle’s braking or stopping distance is affected by: (i) wet roads, (ii) worn-out tyres, (iii) higher vehicle mass, (iv) driving at night, (v) fog, (vi) severe weather (rain, snow, storm), and (vii) driver reaction time. Using this information, design safety posters for your school and prepare a short skit to present it in the assembly.
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Stopping distance increases on wet roads and with worn tyres (less grip), with a heavier vehicle (more momentum), at night and in fog (less visibility, later reaction), in severe weather, and with a slower driver reaction time (longer reaction distance before braking). Posters should urge lower speeds and larger following gaps in these conditions.