# An outsider begging for assistance - motorcycle acceleration

1. Oct 9, 2013

### Jlkohel911

An outsider begging for assistance -- motorcycle acceleration

I have a math problem that, I believe if completed successfully, will save my life.

B_____________________________________________________________________A

The distance from Point A to Point B is 640ft. Starting velocity is 0, ending velocity is 60mph. The bike goes 0-60 in 3.5sec so time is 3.5.

I need to know:
1. Rate of Acceleration (I think it's 7.7meters/sec^2)
2. Distance covered during acceleration (in feet)
3. Distance covered @ 60mph (in feet)

the jist: I was not speeding. The cop said I was and today I'm going to prove it in court. I had 640 ft to get up to 60 miles an hour, go 60mph long enough for him to notice, raise his gun, and clock me, and still come to a safe stop at a red light at point B.

My point is that a 2001 suzuki sv 650s ( the bike i was on at the time) is rated to go 0-60 in 3.5 sec. All the factors of: its not 2001 (so my bike is 12 years old), im not a professional rider, and Im not on a professional track, would to my knowledge only increase distance covered during an acceleration to 60mph. That with a breaking distance of at least 118ft (http://en.wikipedia.org/wiki/Suzuki_SV650) added to distance covered (in ft) per sec while going 60MPH would put me way past 640 ft. However there is some math and conversions that I can't get right on my own and am hoping some of you kind souls might help with. Thank you very much !
Josh

2. Oct 9, 2013

### haruspex

According to my calculations, you only needed 154ft to reach 60mph. That leaves plenty of distance to go somewhat faster, be clocked at that speed, and come to a safe stop.

3. Oct 9, 2013

### arildno

1. Now, 60 mph equals 96.56 kilometres per hour (an hour equals.
2. 640 ft equals 195.07 meters.
3. Now 96.56 km/h=96.56*1000meters/3600 seconds=26,82 m/s
4. Now, dividing that again with 3.5 seconds yields the acceleration 26.82/3.5m/s^2=7,66m/s^2 (as you thought)

5. Now, the distance travelled with this acceleration for 3.5 seconds is:
distance=1/2*acc*time^2=1/2*7.7*3.5^2=47,1625 meters, rather less than 640 feet, I'd say

6

4. Oct 9, 2013

### arildno

Furthermore, using haruspex' feet value, you had left 486 feet to point B, to be traveled at 60mph, i.e 316800 feet ph.

The time to reach B would then be distance/velocity=(486/316800)*3600 seconds=5,5 seconds, or in total from you started, 9 seconds.

5. Oct 9, 2013

### Staff: Mentor

Safe breaking distances for motorcycles can be found online. See for example: BikeSafer.com .

According the the table on that site, safe stopping distance for 60 mph is 210.05 feet*, leaving you with about 430 feet in which to be silly That's more than enough, even if your bike is long in the tooth.

They also tell us that the typical maximum breaking acceleration is -0.774g. From that, and assuming your bike is well maintained and can still do approximately its rated acceleration, you should be able to work out the maximum speed you can hit within that distance. It's a classic intro-physics type problem, so I'll let you work out the details. I'll even tell you that the correct answer is 86.3 mph. Be sure to show your work

*EDIT: The given stopping distances includes a typical reaction time. The specs on your bike give a much more optimistic value of 118.4 feet (Wikipedia article on your bike model). Presumably that value does not include reaction time and assumes perfect conditions.

Last edited: Oct 9, 2013
6. Oct 9, 2013

### Staff: Mentor

After the 154 ft to reach 60, you would be traveling at 88 ft/sec (60 mph). That would leave you 486 ft for him to clock you and for you to come to a stop. If you decelerated at the same rate that you accelerated, this would require 3.5 sec for the deceleration, and another 154 feet to decelerate. This would leave 332 ft for travel at 60. That would occupy about another 3.5 sec. With those acceleration and deceleration rates, that would leave him plenty of time to clock you. Of course, those are pretty high acceleration and deceleration rates (nearly 0.8 g). It isn't clear whether you could achieve those rates without your wheels spinning out. It would take a coefficient of static friction on the order of about 0.8. I don't think the coefficient of static friction of the tire with the road is that high, so your acceleration and deceleration would have had to have been slower. That would have left him less time to clock you. But, even 2 seconds would have been enough.

Chet

7. Oct 9, 2013

### arildno

Our calculations do not prove that you were speeding, though. Is that a comfort?
Neither, for that matter is it proven that your question was not a cleverly represented way to get your homework done by us. Cleverness has its own merits, obviously.

8. Oct 9, 2013

### D H

Staff Emeritus
We don't know what the speed limit was. Apparently the OP got pegged for going 60. I'm assuming that this road is an urban thoroughfare, with a speed limit of 40 or so, 45 tops. (45 is a bit high for a road with stop lights 1/8 mile (660 feet) apart.)

We don't know where the police officer was. Just guessing, was he standing in a parking lot next to his car, ready to pick up errant drivers? If that's the case, suppose the officer heard "vroom. vroom. Varoom!" He looks around and sees a motorcyle first in line at the red light, but raring to go as soon as the light goes green. That officer is going to immediately aim his speed gun at the bike. That's a ticket in the waiting. When the light turns green and the biker doesn't drop out of first gear until he hits 45, the officer knows he has a ticket to issue. The only issue is how big of a ticket.