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Q1.22:
If the speed of light is $3.0 \times 10^8 m s^{-1}$, calculate the distance covered by light in 2.00 ns.

Solution :

Step 1: Given Variables & Conversion to SI Units

To calculate the distance, we must ensure all physical quantities are in base SI units (meters and seconds). The given parameters are:

  • Speed of light ($v$) = $3.0 \times 10^8 \, m \, s^{-1}$
  • Time ($t$) = $2.00 \, ns$

Applying the unit conversion for nanoseconds ($1 \, ns = 10^{-9} \, s$):

$t = 2.00 \times 10^{-9} \, s$ [Definition of the SI prefix 'nano']

Step 2: Selecting the Governing Physical Principle

Since the speed of light in a vacuum is constant, we utilize the fundamental kinematic relationship for uniform motion:

$d = v \times t$ [Per the definition of velocity: $v = \frac{\Delta x}{\Delta t}$]

Step 3: Algebraic Calculation

Substituting the values into the governing equation:

$d = (3.0 \times 10^8 \, m \, s^{-1}) \times (2.00 \times 10^{-9} \, s)$

Perform multiplication of the coefficients and powers of 10 independently:

$d = (3.0 \times 2.00) \times (10^8 \times 10^{-9}) \, m$

$d = 6.0 \times 10^{-1} \, m$

$d = 0.60 \, m$

Visual Representation of Light Propagation

Start (t=0) End (t=2ns) Distance (d) = 0.60 m v = 3.0 × 10⁸ m/s

Step 4: Significant Figures Analysis

In the multiplication $(3.0 \times 10^8) \times (2.00 \times 10^{-9})$:

  • The speed $3.0$ has 2 significant figures.
  • The time $2.00$ has 3 significant figures.
  • Per the rules of significant figures, the final result must be rounded to the least number of significant figures used in the calculation, which is 2.

Our result $0.60 \, m$ correctly maintains 2 significant figures.

Final Solution: The distance covered by light in 2.00 ns is 0.60 m.


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