Question:medium

If the speed of light is 3.0×108 ms–1, calculate the distance covered by light in 2.00 ns.

Updated On: Jan 20, 2026
Show Solution

Solution and Explanation

Given

Speed of light: \[ c = 3.0 \times 10^{8} \ \text{m s}^{-1} \] Time: \[ t = 2.00 \ \text{ns} \]

Step 1: Convert ns to s

\[ 1 \ \text{ns} = 10^{-9} \ \text{s} \] \[ t = 2.00 \ \text{ns} = 2.00 \times 10^{-9} \ \text{s} \]

Step 2: Use Distance Formula

\[ \text{Distance} = \text{Speed} \times \text{Time} \] \[ d = (3.0 \times 10^{8} \ \text{m s}^{-1}) \times (2.00 \times 10^{-9} \ \text{s}) \]

Step 3: Calculate

Multiply coefficients and add exponents: \[ d = 3.0 \times 2.00 \times 10^{8-9} = 6.0 \times 10^{-1} \ \text{m} \] \[ d = 0.60 \ \text{m} \]

Final Answer

The distance covered by light in 2.00 ns is \[ \boxed{0.60 \ \text{m}} \]

Was this answer helpful?
0