AOS4 Topic 10: Rectilinear Motion

Rectilinear motion refers to the motion of an object along a straight line path. In this type of motion, the object moves in one dimension only, typically along a horizontal, vertical, or inclined line, without deviating from its path. This motion can be uniform (constant velocity) or non-uniform (changing velocity).

Vector and Scalar Quantities in Motion

When studying motion, it is important to make a distinction between vector quantities and scalar quantities:

Vector Quantities:

Position, displacement, velocity, and acceleration must be specified by both magnitude and direction.

Scalar Quantities:

Distance, time, and speed are specified by their magnitude only.

Since we are considering movement in a straight line, the direction of each vector quantity is simply specified by the sign of the numerical value.

Position, Velocity, and Acceleration

Position

The position of a particle moving in a straight line is determined by its distance from a fixed point O on the line, called the origin, and whether it is to the right or left of O. By convention, the direction to the right of the origin is considered to be positive.

Consider a particle which starts at \( O \) and begins to move. The position of the particle at any instant can be specified by a real number \( x \). For example, if the unit is meters and if \( x = -3 \), the position is \( 3 \) m to the left of \( O \); while if \( x = 3 \), the position is \( 3 \) m to the right of \( O \). Sometimes there is a rule that enables the position at any instant to be calculated. In this case, we can view \( x \) as being a function of \( t \). Hence \( x(t) \) is the position at time \( t \).

For example, imagine that a stone is dropped from the top of a vertical cliff \( 45 \) meters high. Assume that the stone is a particle traveling in a straight line. Let \( x(t) \) meters be the downwards position of the particle from \( O \), the top of the cliff, \( t \) seconds after the particle is dropped. If air resistance is neglected, then an approximate model for the position is:

\[ x(t) = 5t^2 \quad \text{for} \quad 0 \leq t \leq 3 \]

Displacement and Distance

The displacement of a particle is defined as the change in position of the particle. It is important to distinguish between the scalar quantity distance and the vector quantity displacement (which has a direction). For example, consider a particle that starts at \( O \) and moves first 5 units to the right to point \( P \), and then 7 units to the left to point \( Q \).

The difference between its final position and its initial position is -2. So the displacement of the particle is -2 units. However, the distance it has traveled is 12 units.

Average Velocity

The average rate of change of position with respect to time is average velocity. A particle’s average velocity for a time interval [\( t_1 \), \( t_2 \)] is given by:

average velocity = \(\frac{{\text{{change in position}}}}{{\text{{change in time}}}} = \frac{{x_2 - x_1}}{{t_2 - t_1}}\)

where \( x_1 \) is the position at time \( t_1 \) and \( x_2 \) is the position at time \( t_2 \).

Instantaneous Velocity

The instantaneous rate of change of position with respect to time is instantaneous velocity. We will refer to the instantaneous velocity as simply the velocity.

If a particle’s position, \( x \), at time \( t \) is given as a function of \( t \), then the velocity of the particle at time \( t \) is determined by differentiating the rule for position with respect to time. If \( x \) is the position of a particle at time \( t \), then:

velocity \( v = \frac{{dx}}{{dt}} \)

Note: Velocity is also denoted by \( \dot{x} \) or \( \dot{x}(t) \).

Velocity is a vector quantity. For motion in a straight line, the direction is specified by the sign of the numerical value. If the velocity is positive, the particle is moving to the right, and if it is negative, the particle is moving to the left. A velocity of zero means the particle is instantaneously at rest.

Speed and Average Speed

Speed is a scalar quantity; its value is always non-negative.

  • Speed is the magnitude of the velocity.
  • Average speed for a time interval [\( t_1 \), \( t_2 \)] is given by:
    \(\text{average speed} = \frac{\text{distance travelled}}{t_2 - t_1}\)

Units of Measurement

Common units for velocity (and speed) are:

  • 1 metre per second = 1 m/s = 1 m s-1
  • 1 centimetre per second = 1 cm/s = 1 cm s-1
  • 1 kilometre per hour = 1 km/h = 1 km h-1

The first and third units are connected in the following way:

1 km/h = 1000 m/h = \( \frac{1000}{60 \times 60} \) m/s = \( \frac{5}{18} \) m/s

Therefore, \( 1 \) m/s = \( \frac{18}{5} \) km/h

Example 1

A particle moves in a straight line so that its position, \( x \) cm, relative to \( O \) at time \( t \) seconds is given by \( x = 3t - t^3 \), for \( t \geq 0 \). Find:

  1. a) Its initial position:
  2. b) Its position when \( t = 2 \):
  3. c) Its initial velocity:
  4. d) Its velocity when \( t = 2 \):
  5. e) Its speed when \( t = 2 \):
  6. f) When and where the velocity is zero:

Solution

a) When \( t = 0 \), \( x = 0 \). The particle is initially at \( O \).

b) When \( t = 2 \), \( x = 3 \times 2 - 8 = -2 \). The particle is 2 cm to the left of \( O \).

c) Given \( x = 3t - t^3 \), the velocity is

\[ v = \frac{{dx}}{{dt}} = 3 - 3t^2 \]

When \( t = 0 \), \( v = 3 - 3 \times 0 = 3 \).

The velocity is 3 cm/s. The particle is initially moving to the right.

d) When \( t = 2 \), \( v = 3 - 3 \times 4 = -9 \).

The velocity is -9 cm/s. The particle is moving to the left.

e) When \( t = 2 \), the speed is 9 cm/s. (The speed is the magnitude of the velocity.)

f) \( v = 0 \) implies \( 3 - 3t^2 = 0 \)

\( 3(1 - t^2) = 0 \)

∴ \( t = 1 \) or \( t = -1 \)

But \( t \geq 0 \) and so \( t = 1 \). When \( t = 1 \), \( x = 3 \times 1 - 1 = 2 \).

At time \( t = 1 \) second, the particle is at rest 2 cm to the right of \( O \).

Example 2

A particle moves in a straight line so that its position, \( x \) cm, relative to O at time \( t \) seconds is given by \( x = 3t - t^3 \), for \( t \geq 0 \). Find:

  1. Its initial position
  2. Its position when \( t = 2 \)
  3. Its initial velocity
  4. Its velocity when \( t = 2 \)
  5. Its speed when \( t = 2 \)
  6. When and where the velocity is zero

Solution

  • When \( t = 0 \), \( x = 0 \). The particle is initially at O.
  • When \( t = 2 \), \( x = 3 \times 2 - 8 = -2 \). The particle is 2 cm to the left of O.
  • Given \( x = 3t - t^3 \), the velocity is:

    • \( v = \frac{dx}{dt} = 3 - 3t^2 \)

    When \( t = 0 \), \( v = 3 \). The velocity is 3 cm/s. The particle is initially moving to the right.

  • When \( t = 2 \), \( v = 3 - 3 \times 4 = -9 \). The velocity is -9 cm/s. The particle is moving to the left.

  • When \( t = 2 \), the speed is 9 cm/s. (The speed is the magnitude of the velocity.)
  • \( v = 0 \) implies \( 3 - 3t^2 = 0 \)

    • \( 3(1 - t^2) = 0 \)

    Therefore, \( t = 1 \) or \( t = -1 \). But \( t \geq 0 \) and so \( t = 1 \).

    When \( t = 1 \), \( x = 3 \times 1 - 1 = 2 \).

    At time \( t = 1 \) second, the particle is at rest 2 cm to the right of O.

    Exercise &&1&& (&&1&& Question)

    A particle is moving on x-axis. Its velocity-time graph is given in the figure. Find distance travelled by the particle in 5 seconds.


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