# Thread: Physical Science II - Mph of a dragster at the finish line of a race- ?

1. ## Physical Science II - Mph of a dragster at the finish line of a race- ?

Please assist me with the following question :

One "G" of acceleration is the rate at which an object will accelerate if it is dropped near the Earth's surface, ignoring air resistance. One "G" of force is the force necessary to generate one G of acceleration (although "G" is not an official SI unit of force).

Suppose you built a rocket-powered dragster which could produce a constant acceleration, for a few seconds, that was similar to the acceleration that Cmdr Shepard experienced for the first several minutes of his flight.
On one pass, your dragster produces a constant acceleration of exactly 5.97 G's throughout the whole race.
How fast would your dragster be going at the end of a drag strip that was exactly 1000 feet long?

The work I did so far is:
√2(58.506 m/s˛)(1000m) = 342.07016824 m/s˛ converted to MPH = 765.1892 rounded to 765 MPH

Capture.jpg

2. Originally Posted by student_2018
Please assist me with the following question :

One "G" of acceleration is the rate at which an object will accelerate if it is dropped near the Earth's surface, ignoring air resistance. One "G" of force is the force necessary to generate one G of acceleration (although "G" is not an official SI unit of force).

Suppose you built a rocket-powered dragster which could produce a constant acceleration, for a few seconds, that was similar to the acceleration that Cmdr Shepard experienced for the first several minutes of his flight.
On one pass, your dragster produces a constant acceleration of exactly 5.97 G's throughout the whole race.
How fast would your dragster be going at the end of a drag strip that was exactly 1000 feet long?

The work I did so far is:
√2(58.506 m/s˛)(1000m) = 342.07016824 m/s˛ converted to MPH = 765.1892 rounded to 765 MPH

Capture.jpg
Why did you assume a = 6*g (instead of 5.97*g)? And:

Dx = 1000 ft = 304.8 meters

3. The picture was an example from the book on how to solve the problem. I can't figure it out

4. Originally Posted by student_2018
The picture was an example from the book on how to solve the problem. I can't figure it out
Did you adjust for the length to be converted to meters?

5. Instead of making G into metric units and having to convert the 1000ft into meters use 32ft/sec2 and you would only have to convert ft/sec into miles/hour

vi is 0
vf is what you are looking for
a is g*5.97 or 5.97*32=191 feet per second2 (stop being a commy and use some english units)
d is 1000 feet

vf=sqrt(2*191*1000) feet/second
vf=618 feet/second

Convert to mph

618 feet/second * 1 mile/5280feet * 3600 hour/ 1 second = 421 mph

7. Originally Posted by chris84567
vi is 0
vf is what you are looking for
a is g*5.97 or 5.97*32=191 feet per second2 (stop being a commy and use some english units)
d is 1000 feet

vf=sqrt(2*191*1000) feet/second
vf=618 feet/second

Convert to mph

618 feet/second * 1 mile/5280feet * 3600 hour/ 1 second = 421 mph
1) It's commie

2) You guys are the worst for still using Imperial units, and the OP's teacher is a sadist for assigning a problem with mixed units

3) You can guide posters towards a solution, but don't do people's homework for them on this site; they don't learn any physics or math if you just provide them with the solution.