# Hyperloop : pie in the sky or is it time?

**URL:** <https://boards.straightdope.com/t/hyperloop-pie-in-the-sky-or-is-it-time/666116>\
**Category:** Factual Questions\
**Created:** [August 13, 2013, 6:37am UTC](https://boards.straightdope.com/t/hyperloop-pie-in-the-sky-or-is-it-time/666116 "2013-08-13T06:37:17Z")\
**Posts on this page:** 1\
**Showing post:** 305

<div class="post-metadata">

**Author:** ![Whack-a-Mole](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/whack-a-mole/32/141_2.png) [@Whack-a-Mole](https://boards.straightdope.com/u/Whack-a-Mole)\
**Post date:** [December 23, 2023, 3:52am UTC](https://boards.straightdope.com/t/hyperloop-pie-in-the-sky-or-is-it-time/666116/305 "2023-12-23T03:52:35Z")

</div>

FWIW here is the second question I asked. Note, it had to use the previous question to understand the question.

> How much distance is traveled in that time?

> My apologies, my previous response contained an error. While it’s true that it would take 500 seconds to reach 500 mph with 1g acceleration, the distance traveled during that time wouldn’t simply be 500 miles.
> 
> To calculate the distance traveled, we need to consider the formula for distance under constant acceleration:
> 
> d = vt + 1/2 \* a \* t^2
> 
> where:
> 
> d is the distance traveled  
> v is the initial velocity (in this case, 0 mph)  
> a is the acceleration (1g, which is approximately 32.17 ft/s²)  
> t is the time (500 seconds)
> 
> Plugging in the values, we get:
> 
> d = 0 mph \* 500 seconds + 1/2 \* 32.17 ft/s² \* (500 seconds)²
> 
> d ≈ 402,350 feet
> 
> Converting feet to miles, we get:
> 
> d ≈ 75.42 miles
> 
> Therefore, it would take approximately 500 seconds to reach 500 mph at 1g acceleration, and the distance traveled during that time would be approximately 75.42 miles, not 500 miles. I apologize for the earlier mistake and hope this clarifies the calculations.

---

_[View the full topic](https://boards.straightdope.com/t/hyperloop-pie-in-the-sky-or-is-it-time/666116)._