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# M1 - Constant acceleration watch

1. Hello, I would like some help of these two questions. They are from the MEI Mechanics 1 Constant acceleration, Section 2:Further Examples sheet so if anyone has the answers for the whole sheet where I could work backwards that would be more helpful. Thanks in advance.

4. A train is brought to rest with uniform retardation. It travels 30m in the first 2 seconds, and a further 30m in the next 4 seconds. Find
(i) the initial velocity,
(ii) the retardation,
(iii) total time to come to rest

7. A stone is dropped from the top of a cliff and a second later a 2nd stone is thrown vertically downwards at 15m/s. Both stones reach the bottom at the same time. How high is the cliff?
2. (Original post by Quido)
Hello, I would like some help of these two questions. They are from the MEI Mechanics 1 Constant acceleration, Section 2:Further Examples sheet so if anyone has the answers for the whole sheet where I could work backwards that would be more helpful. Thanks in advance.

4. A train is brought to rest with uniform retardation. It travels 30m in the first 2 seconds, and a further 30m in the next 4 seconds. Find
(i) the initial velocity,
(ii) the retardation,
(iii) total time to come to rest

7. A stone is dropped from the top of a cliff and a second later a 2nd stone is thrown vertically downwards at 15m/s. Both stones reach the bottom at the same time. How high is the cliff?
Ok lets begin with 4

you know that u=u, t=6(total timeof 2+4), s=60(30m+30m) and comes to rest mean v=0
Spoiler:
Show
s=(v+u/2)t where 60=(u+0/2)6 divde by 6 to get 10=u/2 u=20ms^-1
3. for part 2 best use the simplest formula v=u+at
v=0 u=20 t=6 a=a
Spoiler:
Show
0=20+6a deceleration means acceleration will be negative so a= 10/3
4. umm isn't part 3 just 4+2?
5. (Original post by Quido)
Hello, I would like some help of these two questions. They are from the MEI Mechanics 1 Constant acceleration, Section 2:Further Examples sheet so if anyone has the answers for the whole sheet where I could work backwards that would be more helpful. Thanks in advance.

4. A train is brought to rest with uniform retardation. It travels 30m in the first 2 seconds, and a further 30m in the next 4 seconds. Find
(i) the initial velocity,
(ii) the retardation,
(iii) total time to come to rest

7. A stone is dropped from the top of a cliff and a second later a 2nd stone is thrown vertically downwards at 15m/s. Both stones reach the bottom at the same time. How high is the cliff?
defining positive as down we know stone 1 has u=0ms^-1 and stone 2 has u=15ms^-1 Sounds like we need a simultaneous equation
we know a=9.8ms^-2 acceleration due to free fall.
we're gonna use t and s for the formulas DON'T FORGET THAT T=T-1 for the 2nd stone because it's dropped a second later
so s=ut+½at² for both
Spoiler:
Show
stone 1 s=ut+½at² s=0+4.9t²
stone 2 s=ut+½at² s=15(t-1)+4.9(t-1)²
stone2 s=15t-15+4.9(t²-2t+1)
s=15t-15+4.9t²-9.8t+4.9
s=4.9t²5.2t-10.1
so eliminate t by using s=s
4.9t²=4.9t²+5.2t-10.1
5.2t=10.1
t=1.94s(2d.p)
now sub back into s=4.9t²(it's easier than the other equation)
s=4.9(1.94)²
s=18.5m (to 1d.p) remember u used 9.8 for a so you must be consistent and use 1 d.p for your final answer)
6. just wait a second i'm editing xD

dam my head hurts after that xD well it wa sgood practise anyway
7. For these types of questions just use the SUVAT equations.
Write S U V A T at the top of the page and put the corresponding values below the letters. It makes it easy to see what you're doing and what equation you're going to need.
There's only 5 equations you will ever need to do all of these questions.
8. (Original post by B_9710)
For these types of questions just use the SUVAT equations.
Write S U V A T at the top of the page and put the corresponding values below the letters. It makes it easy to see what you're doing and what equation you're going to need.
There's only 5 equations you will ever need to do all of these questions.
2 mins too late buddy xD
also the problem isn't knowing to use SUVAT it's getting the values which is the difficult bit(the mechanics part of the question)
9. (Original post by thefatone)
2 mins too late buddy xD
also the problem isn't knowing to use SUVAT it's getting the values which is the difficult bit(the mechanics part of the question)
This makes no sense. You use the SUVAT equations to find the unknown values. For constant acceleration, and in M1 where energy is not considered in any question, you can only use SUVAT.
10. (Original post by B_9710)
This makes no sense. You use the SUVAT equations to find the unknown values. For constant acceleration, and in M1 where energy is not considered in any question, you can only use SUVAT.
it does makes sense(i just missed a when).
it should've read the problem isn't knowing when to use SUVAT, it's getting the values which is the difficult bit
11. (Original post by thefatone)
Ok lets begin with 4

you know that u=u, t=6(total timeof 2+4), s=60(30m+30m) and comes to rest mean v=0
Spoiler:
Show
s=(v+u/2)t where 60=(u+0/2)6 divde by 6 to get 10=u/2 u=20ms^-1
I understand where you get 20m/s from but when splitting the journey into 2 separate parts you get different deceleration values, I think you get -2.5m/s^2. Also why can't U = 15m/s if you assume that it was travelling at 15/ms for the entire 2 seconds. The question has really confused me.
12. (Original post by Quido)
I understand where you get 20m/s from but when splitting the journey into 2 separate parts you get different deceleration values, I think you get -2.5m/s^2. Also why can't U = 15m/s if you assume that it was travelling at 15/ms for the entire 2 seconds. The question has really confused me.
okay your question states that the deceleration is uniform right?
you can only use suvat when you have constant acceleration.

think about it logically if the object is slowing down(decelerating) then it's going to take more time to travel the same distance the further the object travels right?

you can't use speed=distance/time because you're assuming that acceleration is the same throughout when it's not. The question states that the object is decelerating so since we know that u=20(since we worked it out) u=20 at the very beginning when the object moves but 15m into the motion and the velocity of the object isn't 20ms^-1 anymore right?
13. (Original post by thefatone)
Ok lets begin with 4

you know that u=u, t=6(total timeof 2+4), s=60(30m+30m) and comes to rest mean v=0
Spoiler:
Show
s=(v+u/2)t where 60=(u+0/2)6 divde by 6 to get 10=u/2 u=20ms^-1
Nowhere in the question does it say that the question that the train comes to rest after the 6 seconds. How can you just assume that?
14. (Original post by Quido)
Hello, I would like some help of these two questions. They are from the MEI Mechanics 1 Constant acceleration, Section 2:Further Examples sheet so if anyone has the answers for the whole sheet where I could work backwards that would be more helpful. Thanks in advance.

4. A train is brought to rest with uniform retardation. It travels 30m in the first 2 seconds, and a further 30m in the next 4 seconds. Find
(i) the initial velocity,
(ii) the retardation,
(iii) total time to come to rest

7. A stone is dropped from the top of a cliff and a second later a 2nd stone is thrown vertically downwards at 15m/s. Both stones reach the bottom at the same time. How high is the cliff?
(Original post by B_9710)
Nowhere in the question does it say that the question that the train comes to rest after the 6 seconds. How can you just assume that?
there in the question
15. (Original post by thefatone)
there in the question
If you read part 3, you can see that you cannot make this assumption when dealing with the first two parts

Posted from TSR Mobile
16. (Original post by thefatone)
there in the question
Makes the 3rd question a bit silly then.
17. (Original post by thefatone)
okay your question states that the deceleration is uniform right?
you can only use suvat when you have constant acceleration.

think about it logically if the object is slowing down(decelerating) then it's going to take more time to travel the same distance the further the object travels right?

you can't use speed=distance/time because you're assuming that acceleration is the same throughout when it's not. The question states that the object is decelerating so since we know that u=20(since we worked it out) u=20 at the very beginning when the object moves but 15m into the motion and the velocity of the object isn't 20ms^-1 anymore right?
Hmm, I just tried it again using simultaneous equations.
I used s = ut + 1/2 at^2
I got 30 = 2u - 2a and 60 = 6u - 18a and then solved them to get a = -2.5 and u = 17.5 m/s. And then using v = u + at I found t = 7.
18. (Original post by drandy76)
If you read part 3, you can see that you cannot make this assumption when dealing with the first two parts

Posted from TSR Mobile
That's what I thought. The train could still keep decelerating to rest after the 6 seconds right?
19. (Original post by B_9710)
Makes the 3rd question a bit silly then.
it does but i see no other way than to use time since distance isn't given or mentioned at all in the question.
^^ i take it back
maybe it's worth like a mark or something? or i've don't something horrendously wrong in which case
Zacken TeeEm Student403
i call for your help
20. (Original post by Quido)
Hmm, I just tried it again using simultaneous equations.
I used s = ut + 1/2 at^2
I got 30 = 2u - 2a and 60 = 6u - 18a and then solved them to get a = -2.5 and u = 17.5 m/s.
logical answer you're probably right xD

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