# maths alevel mechanics

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A car travelling on a straight road slows down with a constant deceleration.the car passes a road sign 90 kph and then a post box with speed 36 kph. The distance between the road sign and the post box is 240 metres. find the deceleration of the car in ms-2.

(ps:my teacher is rubbish)

(ps:my teacher is rubbish)

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#4

(Original post by

A car travelling on a straight road slows down with a constant deceleration.the car passes a road sign 90 kph and then a post box with speed 36 kph. The distance between the road sign and the post box is 240 metres. find the deceleration of the car in ms-2.

(ps:my teacher is rubbish)

**Gabriel2000**)A car travelling on a straight road slows down with a constant deceleration.the car passes a road sign 90 kph and then a post box with speed 36 kph. The distance between the road sign and the post box is 240 metres. find the deceleration of the car in ms-2.

(ps:my teacher is rubbish)

Also, please post in the maths forum and you'll get a quick answer (I've moved this thread for you). And please be patient - bumping your thread means there's less chance that someone will see it

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#6

Gabriel, what exam board are you studying for it seems your problem tends to be forgetting the suvat equations or what to use as which substitution. Perhaps if you made some revision cards on the different equations or something similar it may help you remember them as for the question, you are given an initial speed, a final speed and a displacement, you need to find the acceleration so which equation could you use for that?

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(Original post by

Gabriel, what exam board are you studying for it seems your problem tends to be forgetting the suvat equations or what to use as which substitution. Perhaps if you made some revision cards on the different equations or something similar it may help you remember them as for the question, you are given an initial speed, a final speed and a displacement, you need to find the acceleration so which equation could you use for that?

**killerjayko**)Gabriel, what exam board are you studying for it seems your problem tends to be forgetting the suvat equations or what to use as which substitution. Perhaps if you made some revision cards on the different equations or something similar it may help you remember them as for the question, you are given an initial speed, a final speed and a displacement, you need to find the acceleration so which equation could you use for that?

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#8

(Original post by

i have not been introduced to SUVAT as this will be my next lesson,so i want to prepare for it.can you help me please?

**Gabriel2000**)i have not been introduced to SUVAT as this will be my next lesson,so i want to prepare for it.can you help me please?

Try reading a textbook or watch videos that introduce SUVAT e.g. here.

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#9

**Gabriel2000**)

i have not been introduced to SUVAT as this will be my next lesson,so i want to prepare for it.can you help me please?

Okay now for the equations. each one of these equations leaves out one part of SUVAT, meaning that you dont need to use it or know it to find the final answer.

If you are not given S, then you use V=u + at.

If you are not given U, then you use S=vt - (1/2)at^2

If you arent given V, you use S= ut + (1/2)at^2

if you arent given a you use S= (1/2)(u + v)t

and finally if you arent given t, you can use V^2 = U^2 + 2as

It is important to note that when working with acceleration, it can be positive or negative depending on whether the question asks for deceleration or acceleration.

any other questions?

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#10

**Gabriel2000**)

A car travelling on a straight road slows down with a constant deceleration.the car passes a road sign 90 kph and then a post box with speed 36 kph. The distance between the road sign and the post box is 240 metres. find the deceleration of the car in ms-2.

(ps:my teacher is rubbish)

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is this how you work it out.....

u =90

v = 36

s =240

v^2 = u^2 +2as

so.... 36^2 =90^2 +480

36^2 -90^2/480

a = -14.175

u =90

v = 36

s =240

v^2 = u^2 +2as

so.... 36^2 =90^2 +480

36^2 -90^2/480

a = -14.175

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#12

(Original post by

is this how you work it out.....

u =90

v = 36

s =240

v^2 = u^2 +2as

so.... 36^2 =90^2 +480

36^2 -90^2/480

a = -14.175

**Gabriel2000**)is this how you work it out.....

u =90

v = 36

s =240

v^2 = u^2 +2as

so.... 36^2 =90^2 +480

36^2 -90^2/480

a = -14.175

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#13

**Gabriel2000**)

is this how you work it out.....

u =90

v = 36

s =240

v^2 = u^2 +2as

so.... 36^2 =90^2 +480

36^2 -90^2/480

a = -14.175

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#14

(Original post by

It looks like you've got the hang of this. But notice that the distance is given in metres not kilometres but the speeds are KPH. So you should first rewrite the distance to be in kilometres.

**Notnek**)It looks like you've got the hang of this. But notice that the distance is given in metres not kilometres but the speeds are KPH. So you should first rewrite the distance to be in kilometres.

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#15

(Original post by

does that still work? i guess theoretically it does, i do physics so im hardwired to do EVERYTHING in metres and seconds

**killerjayko**)does that still work? i guess theoretically it does, i do physics so im hardwired to do EVERYTHING in metres and seconds

It technically would work but converting everything to metres and seconds is simpler.

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#16

(Original post by

I edited my post after rereading the question

It technically would work but converting everything to metres and seconds is simpler.

**Notnek**)I edited my post after rereading the question

It technically would work but converting everything to metres and seconds is simpler.

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#18

(Original post by

so 90 kph will become 25ms^-

**Gabriel2000**)so 90 kph will become 25ms^-

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10^2=25^2+2a(240)

100-625/480=-1.09 (3sf)

the deaccleration is 1.09

100-625/480=-1.09 (3sf)

the deaccleration is 1.09

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#20

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