Motoring Discussion > Distance travelled - arithmetic Miscellaneous
Thread Author: Crankcase Replies: 7
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 Sat 9 Jul 11 18:48 
 Distance travelled - arithmetic - Crankcase
Daft question. I realised I can't do this sum today as brain isn't working. It can't be as difficult as I'm making it, but I started the conversation with Mrs C and realised I couldn't finish it...

So - car does 0 to 60mph in ten seconds dead. If I start from rest and stop measuring when I get to 60mph, how far will I have travelled?

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 Sat 9 Jul 11 19:19 
 Distance travelled - arithmetic - Focusless
Depends - if you accelerate at a constant rate, then same as 10s of 30mph (60mph is a mile per minute so 10s of 30mph is a 1/12th of that, I think).

But if you take extremes:

1. You accelerate to 60mph immediately and instantaneously, so travel for 10s at 60mph (1/6th of a mile).
2. You only accelerate to 60mph instantaneously after 10s, so don't travel any distance.

Of course neither of these is possible, but it does show that it depends on when you do your acceleration and how much.

Thinking of it as a graph of speed against time, it's the area under the graph, which depends on the shape of the line.
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 Sat 9 Jul 11 21:14 
 Distance travelled - arithmetic - Crankcase
Blimey.

Well it's between nothing and a sixth of a mile then. Ta. At least it isn't obvious!
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 Sat 9 Jul 11 22:32 
 Distance travelled - arithmetic - Dutchie
I ten seconds you have covered 1/6 of a mile is about 500 yards.(I think);)

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 Sat 9 Jul 11 22:12 
 Distance travelled - arithmetic - Dr Prunesqualler
From my O level physics in the 1970's I remember:

s = vt - 0.5 *(at^2)

where s = distance travelled
v= final speed
a = acceleration
t = time

so make the speed into something per second and off you go...

Assumes constant accleration. If not, its back to A level maths and an equation for the rate of change of acceleration (knwn as Jolt or Jerk) and some integration.

Last edited by: Dr Prunesqualler on Sat 9 Jul 11 at 22:12
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 Sat 9 Jul 11 22:21 
 Distance travelled - arithmetic - Number_Cruncher
There's a slightly more direct route, which again, assumes constant acceleration

s = ( (v+u) /2 ) * t

in this case, the inital velocity, u, is zero

s = (v/2) * t

which tallies with Focus's earlier post.


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 Sun 10 Jul 11 01:02 
 Distance travelled - arithmetic - zookeeper
about an inch foot and a thick wide then?
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 Sun 10 Jul 11 12:21 
 Distance travelled - arithmetic - Dutchie
Big Head:)
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