# Calculating Stopping Distance of a Cyclist

• a.mlw.walker
In summary, a cyclist is traveling without pedaling and will eventually stop due to friction. Time readings are taken every 10 meters to calculate acceleration, which is found to be -0.19 meters/second^2. The problem is then worked out using equations for constant deceleration, resulting in a total distance traveled of 6.34 meters. If the cyclist was steering in circles, the number of radians left in their travel before stopping can be determined by dividing the distance by 2πr, where r is the radius of the circle.
a.mlw.walker
Hi,

There is a cyclist rolling but not peddelling, so friction will eventually cause him to stop. You don't know a start speed, but Every 10 metres a time reading is taken to see how long it took him to travel that last ten metres so an acceleraton can be calculated

relative to when readings were taken:
0 - 10m : 2.5seconds
10 - 20m : 3.5 seconds

How do I work out how many metres he will travel?

Can I do a = [(distance/time1)-(distance/time2)]/total time

i.e [10/2.5 - 10/3.5]/(2.5 + 3.5)

= acceleration of -0.19 metres/second^2

Is this correct?

Assuming a constant rate of decleration, then velocity change per unit time is linear.

Those ratios you showed are the average velocity during the two stated intervals.

10 m / 2.5 s = 4.0 m/s

10 m / 3.5 s = 2.86 m/s

How long did it take for this change in average velocity to occur?
(Note it's not the total time).

See if you can figure out from here.

Its quite hard to think about - very easy to make yourself believe the wrong thing when thinking about this.

So, because they are average speeds, that is like saying that is the speed is 5 and 15m into the 20 metre travel.

he is covering 4 m/s 5 metres into the 20m, and 2.86m/s 15m into the 20 metres.

So if i took the average of these two speeds, that is the average number of metres he covers a second, between those two points (between the 5 an 15metre marks),

so his average velocity over the whole 20metres is 3.43 m/s

?

But if this is correct how can i think of it as an acceleration so that i can work out his speed in 50m, or I can work out how far he will travel in total after the first stopwatch is started?

I'm struggling to visualze what could be going on, sorry

If you are having difficulty with the problem then work it out the long way. For instance you know that for a negative acceleration of slowing

X = Xo + Vo*t - 1/2*a*t2

For the first interval since X - Xo is given as 10

10 = Vo1*2.5 - 1/2*a*(2.5)2

For the second interval you know that its Vo2 is the Vo1 of the first -a*t as it slowed over the first 10m. So for the next interval ...

10 = Vo2*t - 1/2*a*t2

which becomes

10 = (Vo1 - 2.5*a)*(3.5) - 1/2*a*(3.5)2

Happily now you have 2 equations and 2 unknowns.

a.mlw.walker said:
You don't know a start speed, but Every 10 metres a time reading is taken to see how long it took him to travel that last ten metres so an acceleraton can be calculated relative to when readings were taken:

0 - 10m : 2.5seconds
10 - 20m : 3.5 seconds

I'm struggling to visualze what could be going on, sorry

I'm not sure about the wording of the problem here. If you're taking readings from a stop watch is the stop watch reset at each interval or allowed to continue running, and starting from a non-zero value? If this is the case, then the unknown reading at 0 m is t0 the time it takes to go 0 to 10m t1 and the time it takes to go 10m to 20m t2.

0 m -> 10 m : reading = t0 + t1 = 2.5 seconds
10 m -> 20 m: reading = t0 + t1 + t2 = 3.5 seconds
and t2 = 1 second

I'll skip the solution for this case for now.

I'll assume the simple case that the stop watch is reset every 10 m then you have
t1 = 2.5 seconds
t2 = 3.5 seconds

Since the decleration is linear, the velocity at the middle of the time interval = the average velocity.
v1 = (10 / 2.5) (m/s) = 4.00 m/s at time 1.25 (0.0 + (2.5 / 2))
v2 = (10 / 3.5) (m/s) = 2.85 m/s at time 4.25 (2.5 + (3.5 / 2))

The change in velocity (4.00 m/s - 2.85 m/s) occurred in 3.00 (4.25-1.25) seconds. What does that tell you about acceleration?

Last edited:
Yeah sorry I meant that the time for each ten metres HAD started again at zero on the stopclock.

The acceleration is (-1.15/3) m/s/s

or 0.383m/s/s

And this is linear. In other words I can say that the last average time recorded was 2.85m/s and the cyclist is slowing at 0.383m/s/s so

constant decceleration, therefore s=uv-0.5at*t

4*2.85 - 0.5*0.383*3*3

therefore the cyclist will travel another 9.68m?

Am I getting there?

If I then said the cyclist was steering in circles, I could divide this by 2PI, and see the number of radians left in his travel before he stops?

You must be on the right track because .38 agrees with the solution made the long way I suggested.

To figure the distance just figure the Vo at the first measurement interval that comes readily from plugging into the first equation namely that

Vo = 10 + 1.25*a = 4 + 1.25(.38) = 4.475 m/s

From there then it's easy to figure the total distance as

(4.475)2 = 2*a*x

where x is the distance from the beginning of the first interval.

That works out for me as 26.34 less the 20m of the 2 intervals or 6.34 m.

Last edited:
Ok, I'll do this again on paper, tomorrow, but am I right, in thinking
"If I then said the cyclist was steering in circles, I could divide this by 2PI, and see the number of radians left in his travel before he stops?"

?

a.mlw.walker said:
Ok, I'll do this again on paper, tomorrow, but am I right, in thinking
"If I then said the cyclist was steering in circles, I could divide this by 2PI, and see the number of radians left in his travel before he stops?"

?

The distance / 2πr is the number of revolutions of the wheel, if that's what you're getting at. But you will need r to determine revolutions or radians.

Jeff Reid said:
I'll assume the simple case that the stop watch is reset every 10 m then you have
t1 = 2.5 seconds
t2 = 3.5 seconds

Since the decleration is linear, the velocity at the middle of the time interval = the average velocity.
v1 = (10 / 2.5) (m/s) = 4.000 m/s at time 1.25 (0.0 + (2.5 / 2))
v2 = (10 / 3.5) (m/s) = 2.857 m/s at time 4.25 (2.5 + (3.5 / 2))

The change in velocity (4.000 m/s - 2.857 m/s) occurred in 3.00 (4.25-1.25) seconds.
acceleration = -1.143 m / (3 s2 ) = - .381 m / s2

a.mlw.walker said:
The acceleration is (-1.15/3) m/s/s or 0.383m/s/s

And this is linear. In other words I can say that the last average time recorded was 2.85m/s and the cyclist is slowing at 0.383m/s/s so

4x2.85 - 0.5 x 0.383 x 32
Where did you get the 3 seconds of time from? How much distance was covered as the bicycle slowed from 4.00 m /s to 2.85 m / s?

acceleration = - .381 m / s2

average velocities
va1 = (10 / 2.5) (m/s) = 4.000 m/s at time 1.25 (0.0 + (2.5 / 2))
va2 = (10 / 3.5) (m/s) = 2.857 m/s at time 4.25 (2.5 + (3.5 / 2))

velocity at the end of last interval:
ve2 = 2.857 m/s + ( - .381 m / s2 ) x (+ 3.5 s / 2) = 2.190 m / s
How much time does it take to stop from 2.190 m / s and how much distance is covered in this time?

or velocity at the beginning of first interval:
vb1 = 4.00 m/s + ( - .381 m / s2 ) x (- 2.5 s / 2) = 4.476m / s
How much time does it take to stop from 4.476 m / s and how much distance is covered in this time?
How much distance past 20 m is coverered in this time?

Last edited:

## 1. How is stopping distance calculated for a cyclist?

The stopping distance of a cyclist can be calculated by adding the reaction distance and braking distance. The reaction distance is the distance a cyclist will travel before they begin to brake after seeing an object in their path. The braking distance is the distance the cyclist will travel while braking to come to a complete stop.

## 2. What factors affect the stopping distance of a cyclist?

The stopping distance of a cyclist can be affected by the speed of the cyclist, the condition of their brakes, the surface of the road, and their reaction time. Other factors such as weather conditions and the weight of the cyclist and their bike can also play a role.

## 3. How can I improve my stopping distance as a cyclist?

To improve your stopping distance as a cyclist, make sure to maintain your bike and its brakes in good condition. Also, practice good reaction time by staying alert and anticipating potential hazards. Adjust your speed according to the road and weather conditions to give yourself enough time to react and brake.

## 4. Is there a formula for calculating stopping distance of a cyclist?

Yes, the formula for calculating stopping distance of a cyclist is: Stopping Distance = Reaction Distance + Braking Distance. The reaction distance can be calculated by multiplying the speed of the cyclist by their reaction time. The braking distance can be calculated by using the formula (speed^2)/(2 x deceleration), where deceleration is the rate at which the cyclist's speed decreases while braking.

## 5. How does the stopping distance of a cyclist compare to that of a car?

The stopping distance of a cyclist is typically shorter than that of a car, as cyclists are able to react and stop more quickly due to their smaller size and maneuverability. However, this may vary depending on the speed and conditions of both the cyclist and the car. It is important for both cyclists and drivers to practice safe and cautious behavior on the road to avoid accidents.

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