The Student Room Group

m1 question

For part i) the max speed is reached when t is greater than 4 so why do we sub in t=4 into v= 4t - (1/2)t^2 to get v=8 which is the max speed? doesnt really make sense
help please
hello.PNG
Reply 1
Original post by h26
For part i) the max speed is reached when t is greater than 4 so why do we sub in t=4 into v= 4t - (1/2)t^2 to get v=8 which is the max speed? doesnt really make sense
help please
hello.PNG


Because she reaches max speed at t = 4
Reply 2
Original post by M4cc4n4
Because she reaches max speed at t = 4

Thanks but how do you know she reaches max speed at t=4?
Reply 3
Can you show how you have approached this question?
Reply 4
Original post by h26
Thanks but how do you know she reaches max speed at t=4?


Because she stops accelerating at t=4
Reply 5
Original post by h26
Thanks but how do you know she reaches max speed at t=4?


At t=4, her acceleration goes to zero so speed doesn't change therefore it will be and stay max
Reply 6
max speed is at t=4 because the function for acceleration is zero at that point and onwards.
Reply 7
Original post by ThomH97
Can you show how you have approached this question?

So whenever the acceleration is zero, then max speed has been reached

the acceleration is zero here when t is greater than 4.
The inequality sign is confusing me because she doesn't really stop accelerating at t=4
Reply 8
Original post by Joshey
At t=4, her acceleration goes to zero so speed doesn't change therefore it will be and stay max


But it says in teh question a=0 for t>4 -it doesnt even include t=4
Reply 9
Original post by sstyan
max speed is at t=4 because the function for acceleration is zero at that point and onwards.


But it says in teh question a=0 for t>4 -it doesnt even include t=4
Reply 10
Original post by M4cc4n4
Because she stops accelerating at t=4


But it says in teh question a=0 for t>4 -it doesnt even include t=4
Reply 11
Original post by h26
But it says in teh question a=0 for t>4 -it doesnt even include t=4


It's a convention thing and you're looking too deeply at it. You can always safely assume what we said as true. I understand I was confused by this when I first took physics
Original post by h26
So whenever the acceleration is zero, then max speed has been reached

the acceleration is zero here when t is greater than 4.
The inequality sign is confusing me because she doesn't really stop accelerating at t=4


It means that for all time after 4 seconds, she doesn't accelerate anymore. You could get a question where she stopped accelerating at 4 seconds, maintained her speed for a while then slowed down to a stop. This would look something like:

a = 4 - t, 0 < t < 4
a = 0, 4 < t < 10
a = -2, 10 < t < 14
a = 0, 14 < t
Reply 13
Original post by Joshey
It's a convention thing and you're looking too deeply at it. You can always safely assume what we said as true. I understand I was confused by this when I first took physics


So I guess you just have to accept that if the previous inequality has e.g. 4 in it at the end and the other inequality (which if for a =0)has a 4 in it at the end then 0 acceleration is when t=4 regardless of what the inequality sign is? Is that correct?
Reply 14
Original post by ThomH97
It means that for all time after 4 seconds, she doesn't accelerate anymore. You could get a question where she stopped accelerating at 4 seconds, maintained her speed for a while then slowed down to a stop. This would look something like:

a = 4 - t, 0 < t < 4
a = 0, 4 < t < 10
a = -2, 10 < t < 14
a = 0, 14 < t

So I guess you just have to accept that if the previous inequality has e.g. 4 in it at the end and the other inequality (which if for a =0)has a 4 in it at the end then 0 acceleration is when t=4 regardless of what the inequality sign is? Is that correct?
Original post by h26
So I guess you just have to accept that if the previous inequality has e.g. 4 in it at the end and the other inequality (which if for a =0)has a 4 in it at the end then 0 acceleration is when t=4 regardless of what the inequality sign is? Is that correct?


I'm not sure what you are asking. The two lines there are telling you the acceleration in two phases of motion. The first being between starting and 4 seconds. The second being after 4 seconds for some unknown time (which you then find).
Reply 16
Original post by ThomH97
I'm not sure what you are asking. The two lines there are telling you the acceleration in two phases of motion. The first being between starting and 4 seconds. The second being after 4 seconds for some unknown time (which you then find).

No worries I got it thank you very much:smile:
Reply 17
Original post by h26
So I guess you just have to accept that if the previous inequality has e.g. 4 in it at the end and the other inequality (which if for a =0)has a 4 in it at the end then 0 acceleration is when t=4 regardless of what the inequality sign is? Is that correct?


Pretty much. a = 0 at ~t and above it's fine to use in this case

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