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
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
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
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
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?
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?
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).
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).
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