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GCSE physics stopping distance question! please help!

OK I won't bother going through the whole question but basically:

Car travelling at 15 m/s. Thinking distance = 9m, Braking distance = 19m so overall stopping distance = 28m.
-->
Speed = Distance/Time, therefore thinking time = 0.6 secs and braking time = 1.27 secs.
-->
Mass of car = 1000kg.
-->
Ke = 1/2mv^2, therefore Ke of car = 112500 J.
-->
SHOW THAT THE BRAKING FORCE NEEDED TO STOP THE CAR IN 19M IS ABOUT 5900 METRES.

Cheers, S.C
I'll assume you mean 5900N

u = 15m/s, s = 19m, a = ?, v = 0m/s
= + 2as
0 = 15² + 2(19)a
-38a = 225
a = -5.92m/s²

-F = ma
-F = 1000 x -5.92 = -5921
F = 5921N

oops, just realised it's gcse.
StriatedCentipe
OK I won't bother going through the whole question but basically:

Car travelling at 15 m/s. Thinking distance = 9m, Braking distance = 19m so overall stopping distance = 28m.
-->
Speed = Distance/Time, therefore thinking time = 0.6 secs and braking time = 1.27 secs.
-->
Mass of car = 1000kg.
-->
Ke = 1/2mv^2, therefore Ke of car = 112500 J.
-->
SHOW THAT THE BRAKING FORCE NEEDED TO STOP THE CAR IN 19M IS ABOUT 5900 METRES.

Cheers, S.C


Equate KE to Work done by brakes

work = force * distance

ie KE = F*d

plug in your numbers and there you go :smile:

Edit: and just to prove how great physics is it gives exactly the same answer as Widowmaker but in 1 line :smile: . Just shows energy conservation and kinematics give exactly the same answer - as it should. :smile:
Reply 3
no it's alright I can deal with UVAST :P
StriatedCentipe
no it's alright I can deal with UVAST :P


erm... but why.. if I had the choice of 1 line to 10 I choose the shortest every time. Energy conservation is a more elegant method. Even at degree I use energy conservation in preference to SUVAT. In fact I am in the process of doing just such a question :p: :rofl:
Reply 5
thanks both!
Reply 6
a kind of follow-up question:

'When the speed is doubled, the overall stopping distance is more than doubled. Explain why.'

I said
v²=u²+2as
u=0 so ignore
v²=2as
a is same, even when speed is doubled so ignore
v²=2s
so s=v²/2
2 is constant
s=v²
therefore, as v increases in a linear fashion, s increases in a non-linear fashion.

...Can I say this? I make some pretty dumb mistakes :redface:
StriatedCentipe
a kind of follow-up question:

'When the speed is doubled, the overall stopping distance is more than doubled. Explain why.'

I said
v²=u²+2as
u=0 so ignore
v²=2as
a is same, even when speed is doubled so ignore
v²=2s
so s=v²/2
2 is constant
s=v²
therefore, as v increases in a linear fashion, s increases in a non-linear fashion.

...Can I say this? I make some pretty dumb mistakes :redface:


erm yeah you can to an extent. you cant just get rid of the a though... its still there its just constant. And instead of non-linear how about doubling speed quadruples distance - ie its quadratic :smile:
= 2as
so at first speed v, s = v²/2a
at doubled speed 2v, s = (2v)²/2a = 4v²/2a
Reply 9
Widowmaker
= 2as
so at first speed v, s = v²/2a
at doubled speed 2v, s = (2v)²/2a = 4v²/2a


oh ok thanks. This qu was about in the middle of a past gcse paper, and what I did isn't on the syllabus, so is there another way of expressing this without using eqns of motion?
StriatedCentipe
oh ok thanks. This qu was about in the middle of a past gcse paper, and what I did isn't on the syllabus, so is there another way of expressing this without using eqns of motion?


yeah using energy like I did:

In this case equate KE and work done:

(1/2)mv^2 = F*d
d=[(1/2)m/F]*v^2

d is proportional to v^2

hence the advertised result :smile:
Reply 11
F1 fanatic
yeah using energy like I did:

In this case equate KE and work done:

(1/2)mv^2 = F*d
d=[(1/2)m/F]*v^2

d is proportional to v^2

hence the advertised result :smile:


thanks dude!