physics suvat question help

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

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)

Reply 2

helpicantgetup

9

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

Reply 3

RogerOxon

21

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 :smile:

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}

Reply 4

RogerOxon

21

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.

(edited 5 years ago)

Original post by RogerOxon

I'd refuse to answer. You don't ride a car :smile:

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?

Reply 6

RogerOxon

21

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}

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

Reply 8

helpicantgetup

9

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!

Reply 9

RogerOxon

21

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