Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A lightyear is defined as the distance that light travels in a year

(i) calculate, in meters, the distance of $1$ light year, give your answer In standard form( speed of light=$3\times10^8$ m/s, numbers of days in a year = $365$

(ii) the nearest star from earth, alpha Centauri, is approximately $4.24$ light years away. Calculate in years, the time taken for a spaceship traveling at $1.2\times10^5$ m/s to reach the nearest star

share|improve this question
What have you tried so far? –  Sujaan Kunalan Jun 17 '13 at 19:04
I did (i) but I'm stuck at(ii) –  user82844 Jun 17 '13 at 19:14
That's not the speed of light. The speed of light is $2.99792458\times 10^8 \mathrm m/\mathrm s$, exactly. –  dfeuer Jun 17 '13 at 19:17
Some questions round it to 3 $\times$ 10^8 m/s –  Sujaan Kunalan Jun 17 '13 at 19:18
That's why we have the $\approx$ symbol. –  dfeuer Jun 17 '13 at 19:18
add comment

2 Answers

up vote 0 down vote accepted

For part (i), you should use the relation $\text{speed}=\dfrac{\text{distance}}{\text{time}}$ and solve for $\text{distance}$ where $\text{speed}=3\times10^8\text{m/s}$ and $\text{time}=1\times 365\times24\times60\times60 \text{s}$.

For part (ii), once you've figured out what a light year is in metres in part (i), you can again use the relation $\text{speed}=\dfrac{\text{distance}}{\text{time}}$ and solve for time where $\text{distance}=4.24\times\text{light years in metres}$, and $\text{speed}=1.2\times10^5\text{m/s}$. Once you've obtained your time in seconds, convert it into years.

share|improve this answer
add comment

Hint 1: For (i), you are given the speed and the time. Given the formula: time $\times$ speed = distance, can you solve it now?

Hint 2: For part (ii), convert light years into metres, and then apply the formula listed above.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.