This chapter develops a deeper understanding of motion through observation and measurement. It introduces frame of reference, relative motion, scalar and vector quantities, graphical vector addition, and equations of motion for uniformly accelerated motion.
Key ideas: Frame of reference • Relative motion • Scalars and vectors • Vector addition • Equations of motion
Motion is described by observing a change in an object's position with time relative to a reference point. Motion may be slow or fast, straight or curved, uniform or non-uniform.
Speed tells us how quickly distance is covered:
Speed = Distance ÷ Time
A frame of reference is the reference point or coordinate system with respect to which the position and motion of an object are described. A frame at rest or moving with constant velocity is an inertial frame; an accelerating frame is non-inertial.
A scalar has magnitude only. Examples include distance, time, mass, speed and work.
A vector has both magnitude and direction. Examples include displacement, velocity and force.
To add vectors by the triangle method, place the tail of the second vector at the head of the first vector. The vector joining the tail of the first to the head of the second is the resultant.
For motion with constant acceleration:
v = u + at
s = ut + ½at2
v2 = u2 + 2as
The distance travelled in the nth second is:
sn = u + ½a(2n − 1)
Here, u is initial velocity, v is final velocity, a is acceleration, t is time and s is displacement.
The chapter gives the coordinates A(1,1), B(3,1), C(3,5) and D(4,5), representing Sita's house, bus stop, traffic signal and school respectively. The coordinates are in kilometres.
She walks from A to B.
AB = 3 − 1 = 2 km.
The bus travels from B to C and then C to D.
BC = 5 − 1 = 4 km and CD = 4 − 3 = 1 km.
Therefore, bus distance = 4 + 1 = 5 km.
Displacement is the straight-line distance from A to D.
AD = √[(4 − 1)2 + (5 − 1)2]
= √(9 + 16) = √25 = 5 km.
Answer: A frame of reference is a reference point or coordinate system with respect to which we describe the position and motion of an object. Without specifying it, we cannot decide whether an object is at rest or in motion.
Answer:
Answer: Both passengers have the same position relative to each other and move together with the same velocity. Therefore, the relative position between them does not change, so one passenger appears at rest to the other.
Answer:
Answer: Distance is the total path length travelled, so it is a scalar. Displacement is the shortest straight-line change in position from the starting point to the final point, so it is a vector.
For the activity in the chapter, walk from A to B and then return from B to A. The total distance is the complete path travelled, but the final position is the same as the starting position. Therefore, displacement is zero.
Simple representation: A → B → A. The travelled path is non-zero, while the final displacement from A to A is 0.
Answer: Velocity of a moving car and force applied while pushing a box are two everyday examples of vector quantities because both require magnitude and direction.
Solution: Let the first vector be 4 units east and the second be 3 units north. The vectors are perpendicular.
Using the right-triangle relation:
R = √(42 + 32)
= √(16 + 9) = √25 = 5 units.
The resultant is directed towards the north-east. Its angle with the east direction is tan−1(3/4) ≈ 36.9°.
Answer: Subtracting vector B from vector A is equivalent to adding the negative of B:
A − B = A + (−B).
To draw −B, keep the magnitude of B unchanged but reverse its direction. Then add A and −B by the triangle or parallelogram method. The resultant gives A − B.
Solution: Let one vector be 5 units east and the other be 5 units west.
Taking east as positive:
R = +5 + (−5) = 0.
Thus, two equal vectors acting in opposite directions have a zero resultant.
Given: u = 0, a = 4 m/s2, n = 6.
Distance travelled in the nth second:
sn = u + ½a(2n − 1)
s6 = 0 + ½(4)(2 × 6 − 1)
= 2 × 11 = 22 m.
Given: u = 8 m/s, a = 2 m/s2, n = 5.
sn = u + ½a(2n − 1)
s5 = 8 + ½(2)(2 × 5 − 1)
= 8 + 9 = 17 m.
Motion and rest are relative. An object's position must be compared with a reference point or frame before we can say whether its position is changing.
The magnitude tells how far the final position is from the initial position, while direction tells where the final position lies relative to the initial position. Together they completely specify displacement.
Starting a car, a bicycle speeding up, a bus braking, a lift starting upward or downward, and a ball thrown upward are examples in which velocity changes with time.
Choose a straight path and mark two points a known distance apart. Measure the time taken by a walking student or cyclist to travel between them. Calculate speed using speed = distance/time. Repeat the measurement several times, record the observations in a table, calculate the speeds and compare the results. A conclusion should mention possible sources of experimental error such as reaction time and inaccurate distance measurement.