The Drag Race

deanx0r

Senior member
Oct 1, 2002
890
20
76
Someone brought this problem to me the other day which I found rather interesting because it doesn't require extensive maths or physics know-how to be solved. Answer will be posted by friday 15th.

James and Harold are drag racing from a standing start and at a constant acceleration.

James covers the last quarter of the distance in 3 seconds.

Harold covers the last third of the distance in 4 seconds.

Who wins the race, and by how much?

Hint for those without the physics background: start with the following kinematic equation.

d= v0 + (1/2)at^2

where d is the distance, v0 is the initial speed, t is time and a is acceleration.
 
Sep 12, 2004
16,852
59
86
run-baby-run.jpg


Which one is James and which one is Harold?
 

phucheneh

Diamond Member
Jun 30, 2012
7,306
5
0
Cars do not accelerate at a set pace. This is not how drag racing works.

Give us their ET's and trap speeds. ;P
 

Kadarin

Lifer
Nov 23, 2001
44,296
16
81
Cars do not accelerate at a set pace. This is not how drag racing works.

Give us their ET's and trap speeds. ;P

This. Also need 60ft time and reaction time to find out who got the holeshot. The faster car doesn't necessarily win the race.
 

gorobei

Diamond Member
Jan 7, 2007
3,846
1,289
136
im told the car with flames painted on the side and spinner hubs are the fastest.
 

3chordcharlie

Diamond Member
Mar 30, 2004
9,859
1
81
im told the car with flames painted on the side and spinner hubs are the fastest.

Not if the other car has white faced gauges, preferably with fuel and oil pressure mounted on the driver's A-pillar.

Muffler-back exhaust upgrades help a lot too.
 

Vdubchaos

Lifer
Nov 11, 2009
10,408
10
0
Someone brought this problem to me the other day which I found rather interesting because it doesn't require extensive maths or physics know-how to be solved. Answer will be posted by friday 15th.

James and Harold are drag racing from a standing start and at a constant acceleration.

James covers the last quarter of the distance in 3 seconds.

Harold covers the last third of the distance in 4 seconds.

Who wins the race, and by how much?

Hint for those without the physics background: start with the following kinematic equation.

d= v0 + (1/2)at^2

where d is the distance, v0 is the initial speed, t is time and a is acceleration.

It really depends how much distance they covered previous to last quarter and third.

Without knowing those #s, it would be impossible to get a resolution.
 

KMc

Golden Member
Jan 26, 2007
1,149
0
76
Ok, call me crazy, but if they are both at a standing start (v0=0) and they both are accelerating at the same rate (a) then at any time (t), they will both have covered the same distance (d). So all other information is irrelevant - the race is a tie.
 

Crusty

Lifer
Sep 30, 2001
12,684
2
81
Ok, call me crazy, but if they are both at a standing start (v0=0) and they both are accelerating at the same rate (a) then at any time (t), they will both have covered the same distance (d). So all other information is irrelevant - the race is a tie.

Either that or each car accelerates at a different constant rate.
 

deanx0r

Senior member
Oct 1, 2002
890
20
76
Ok, call me crazy, but if they are both at a standing start (v0=0) and they both are accelerating at the same rate (a) then at any time (t), they will both have covered the same distance (d). So all other information is irrelevant - the race is a tie.

The acceleration is constant, but they both accelerate at a different rate.

It really depends how much distance they covered previous to last quarter and third.

Without knowing those #s, it would be impossible to get a resolution.

Hint for those trying to set this up:

You can split up the race in two parts. For James' race, you know he covered the first 3/4 distance with an initial velocity of 0 (v0). And the last 1/4 distance was covered in 3 seconds.


|_________3/4d_______________|____1/4d____|

From d= v0t + (1/2)at^2 you then have
3/4d = 1/2at^2 + v0t

Solving for d , you have:
d = 2/3at^2

Now for the last 1/4 of the race.
He covered the last 1/4 in 3 seconds.

1/4d = 1/2a(3)^2 + v(3)
Velocity can also be expressed as v = at where t is the time at the 3/4 mark.
1/4d = 1/2a(3)^2 + at(3)

Solving for d , you have:
d = 18a+12at

It's almost there, right?
 

DesiPower

Lifer
Nov 22, 2008
15,299
740
126
by my rough estimate, I think the second dude won.

Even at constant velocity, the 2nd dude covered the last 1/4 in 2.99 second, but he was accelerating so the time it took him to cover the .0825 of the distance was definitely more than the time to cover the last .25 of the distance. so the 2nd dude covered the last .25 of the distance in less than 2.99 seconds... as the acceleration was constant, he definitely was faster in the last .25 of the distance. So my Vote - Dude #2
 
Last edited:

KMc

Golden Member
Jan 26, 2007
1,149
0
76
I calculate Harold wins with a run of 21.80 seconds.
James finishes in 22.38 seconds.
 

BlitzPuppet

Platinum Member
Feb 4, 2012
2,460
7
81
Gay. Powerbands anyone? What if one is all low end acceleration and the other doesn't pick up until 40mph?
 

deanx0r

Senior member
Oct 1, 2002
890
20
76
I calculate Harold wins with a run of 21.80 seconds.
James finishes in 22.38 seconds.

Yup, that is the correct answer. Good job.

I am posting the solution below.


gif.latex


gif.latex


the remainder quarter of the race can can be written as followed:

gif.latex


gif.latex


gif.latex


gif.latex


Set them both equal to each other to solve for t:

gif.latex


gif.latex


gif.latex
 

deanx0r

Senior member
Oct 1, 2002
890
20
76
gif.latex


gif.latex


Total time for James:

gif.latex

Total time for Harold:

gif.latex


gif.latex


Harolds wins by 0.594s.
 
Last edited: