# physics suvat question help

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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

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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#2

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)

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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#3

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

49 - 36 = 13m

Unless you mean something different

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#4

(Original post by

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)?

**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 consider the time, , to start at t=5, so put (from ) and into this:

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#5

(Original post by

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

**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

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

And calculate the difference. That works, as does the method that I outlined earlier.

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#6

(Original post by

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

I'd consider the time, , to start at t=5, so put (from ) and into this:

**RogerOxon**)I'd refuse to answer. You don't ride a car

I'd consider the time, , to start at t=5, so put (from ) and into this:

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#7

(Original post by

why is intial velocity 10?

**Gent2324**)why is intial velocity 10?

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#8

(Original post by

At t=5, the car has been accelerating at a raet of for 5s, so is going at

**RogerOxon**)At t=5, the car has been accelerating at a raet of for 5s, so is going at

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#9

(Original post by

The sixth second is that between t=5 and 6, not 6 and 7.

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

And calculate the difference. That works, as does the method that I outlined earlier.

**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:

And calculate the difference. That works, as does the method that I outlined earlier.

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#10

(Original post by

oh i see, so was my answer right or completely wrong

**Gent2324**)oh i see, so was my answer right or completely wrong

At t=5, the car is travelling at .

At t=6, the car is travelling at .

yet another way to do this is to calculate the distance traveled from the average speed over that second.

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(Original post by

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?

**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?

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