Question.
If the speed of light is $3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$, calculate the distance covered by light in $2.00 \mathrm{~ns}$.
If the speed of light is $3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$, calculate the distance covered by light in $2.00 \mathrm{~ns}$.
Solution:
According to the question:
Time taken to cover the distance $=2.00 \mathrm{~ns}$
$=2.00 \times 10^{-9} \mathrm{~s}$
Speed of light $=3.0 \times 10^{8} \mathrm{~ms}^{-1}$
Distance travelled by light in $2.00 \mathrm{~ns}$
$=$ Speed of light $\times$ Time taken
$=\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\left(2.00 \times 10^{-9} \mathrm{~s}\right)$
$=6.00 \times 10^{-1} \mathrm{~m}$
$=0.600 \mathrm{~m}$
According to the question:
Time taken to cover the distance $=2.00 \mathrm{~ns}$
$=2.00 \times 10^{-9} \mathrm{~s}$
Speed of light $=3.0 \times 10^{8} \mathrm{~ms}^{-1}$
Distance travelled by light in $2.00 \mathrm{~ns}$
$=$ Speed of light $\times$ Time taken
$=\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)\left(2.00 \times 10^{-9} \mathrm{~s}\right)$
$=6.00 \times 10^{-1} \mathrm{~m}$
$=0.600 \mathrm{~m}$
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