maths alevel mechanics
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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anyone......please
Original post by 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)
(ps:my teacher is rubbish)
You've got the initial speed (u), the final speed (v) and the displacement (s). Any thoughts?
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
???????
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?
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?Original post by 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?
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?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?
Original post by 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?
If you don't know the topic then you need to learn if before trying a question
Try reading a textbook or watch videos that introduce SUVAT e.g. here.
Original post by 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?
of course, in a "SUVAT" equation there are (obviously) 5 different parts. S is used to represent displacement, dont get it confused with distance though they are different things (do you already know that). U is the "initial velocity" which is the speed of the object at the start of the question you could say, so if an object is being dropped U would be 0, or say somebody throws a ball at 5ms^-1, and the question asks you to calculate something after it leaves their hand then u would be 5. V is the final velocity, which is the velocity that the object reaches at the end of its journey, if a question says the object comes to rest then V will be equal to 0, if not it will tell you what the value is or you will have to calculate it. As im sure you can guess, a stands for acceleration, which again will either be given to you, youd have to assume it e.g. 9.8 for falling objects, or youd have to calculate it either using SUVAT or using Forces. and finally, T stands for time, which obviously is just the amount of time that the object travels for.
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?
Original post by 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)
(ps:my teacher is rubbish)
you could always draw a v-t graph, the gradient is the acceleration and the area under the graph is the distance
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
Original post by 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
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
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 change the units first.
(edited 6 years ago)
Original post by 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
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
almost, youd have to do the conversi9ns first from kph to ms^-1. i forgot to mention the units. in all of mechanics, unless it states otherwise, you should ALWAYS use ms^-1, m, ms^-2 and s. so to convert from kph to ms^- simply multiple by 5/18
Original post by 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.
does that still work? i guess theoretically it does, i do physics so im hardwired to do EVERYTHING in metres and seconds
Original post by killerjayko
does that still work? i guess theoretically it does, i do physics so im hardwired to do EVERYTHING in metres and seconds
I edited my post after rereading the question
It technically would work but converting everything to metres and seconds is simpler.
Original post by Notnek
I edited my post after rereading the question
It technically would work but converting everything to metres and seconds is simpler.
It technically would work but converting everything to metres and seconds is simpler.
okay thought so
SI units are bae <3so 90 kph will become 25ms^-
Original post by Gabriel2000
so 90 kph will become 25ms^-
Yes that's right. Do the same for the other speed then go through the same working again and you should get the right answer.
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
Original post by Gabriel2000
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
This looks good. If your teacher isn't great then please continue to post in the maths forum whenever you are stuck. But make sure you post maths questions here and not anywhere else on TSR
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