# Stopping distance

#### a.mlw.walker

Hi,

There is a cyclist rolling but not peddelling, so friction will eventually cause him to stop. You dont 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???

#### rcgldr

Homework Helper
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.

#### a.mlw.walker

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

#### LowlyPion

Homework Helper
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.

#### rcgldr

Homework Helper
You dont 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) occured in 3.00 (4.25-1.25) seconds. What does that tell you about acceleration?

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#### a.mlw.walker

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?

#### LowlyPion

Homework Helper
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.

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#### a.mlw.walker

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?"

???

#### LowlyPion

Homework Helper
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.

#### rcgldr

Homework Helper
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) occured in 3.00 (4.25-1.25) seconds.
acceleration = -1.143 m / (3 s2 ) = - .381 m / s2

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?

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