x_datboi_x
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a car accelerates uniformly at 2 m/s^2 for 10 seconds. if the rider starts from rest ,work out the distance travelled in the sixth second(between 5 and 6)?

I have worked out the distance travelled in the 10 seconds using s=ut+0.5at^2 but idk what the distance is after at the 6th second ( it's not 36)

please show how you got your answer thanks
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Gent2324
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#2
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S
U 0
V 20
A 2
T 10

we need to work out the velocity at 6 seconds, so V= u + at 0+2x10 = 20
using s = vt - 1/2at^2 i got 84, not sure if its right though we'd need some physics person to help.
(tag me in the real answer please)
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helpicantgetup
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Why not just work out the distance for 6seconds and then the distance for 7 seconds then do the 7 sec - 6 sec = distance travelled in the sixth second? In this case it would be
49 - 36 = 13m
Unless you mean something different
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RogerOxon
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(Original post by Here_comedatboi)
a car accelerates uniformly at 2 m/s^2 for 10 seconds. if the rider starts from rest ,work out the distance travelled in the sixth second(between 5 and 6)?
I'd refuse to answer. You don't ride a car

I'd consider the time, t', to start at t=5, so put u=10 ms^{-1} (from v=u+at) and t'=1 into this:
s=ut'+\frac{a{t'}^2}{2}
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RogerOxon
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(Original post by helpicantgetup)
Why not just work out the distance for 6seconds and then the distance for 7 seconds then do the 7 sec - 6 sec = distance travelled in the sixth second? In this case it would be
49 - 36 = 13m
Unless you mean something different
The sixth second is that between t=5 and 6, not 6 and 7.

If you want to this approach, then you'd do:

s=ut+\frac{at^2}{2}=\frac{at^2}{2}

s_{t=5}=\frac{at^2}{2}=..
s_{t=6}=\frac{at^2}{2}=..

And calculate the difference. That works, as does the method that I outlined earlier.
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Gent2324
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(Original post by RogerOxon)
I'd refuse to answer. You don't ride a car

I'd consider the time, t', to start at t=5, so put u=10 ms^{-1} (from v=u+at) and t'=1 into this:
s=ut'+\frac{a{t'}^2}{2}
why is intial velocity 10?
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RogerOxon
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(Original post by Gent2324)
why is intial velocity 10?
At t=5, the car has been accelerating at a raet of 2 ms^{-2} for 5s, so is going at 10 ms^{-1}
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Gent2324
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(Original post by RogerOxon)
At t=5, the car has been accelerating at a raet of 2 ms^{-2} for 5s, so is going at 10 ms^{-1}
oh i see, so was my answer right or completely wrong
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helpicantgetup
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(Original post by RogerOxon)
The sixth second is that between t=5 and 6, not 6 and 7.

If you want to this approach, then you'd do:

s=ut+\frac{at^2}{2}=\frac{at^2}{2}

s_{t=5}=\frac{at^2}{2}=..
s_{t=6}=\frac{at^2}{2}=..

And calculate the difference. That works, as does the method that I outlined earlier.
Oh yep, I forgot to factor in the fact that the 1st second starts from zero and so on, thanks!
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RogerOxon
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(Original post by Gent2324)
oh i see, so was my answer right or completely wrong
84 isn't the correct answer. I won't give the exact one, as the OP needs to understand how to do the questions, but it's not too far off 10m.

At t=5, the car is travelling at 10 ms^{-1}.
At t=6, the car is travelling at 12 ms^{-1}.

yet another way to do this is to calculate the distance traveled from the average speed over that second.
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ibyghee
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#11
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s=1/2 at^2
s=1/2*2*5^2 =25
s=1/2*2*6^2=36
36-25=11? he travelled 11 meteres no?
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x_datboi_x
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(Original post by ibyghee)
s=1/2 at^2
s=1/2*2*5^2 =25
s=1/2*2*6^2=36
36-25=11? he travelled 11 meteres no?
thanks, it is 11
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