Original post by 2014_GCSE
I feel like it's easy but I just can't do it for some reason. What am I missing?
I've tried:
If s = 15, but you can't sub t in so how do you find out A?
If s = 0, t = 0, and you can't find A here.
I feel like it's easy but I just can't do it for some reason. What am I missing?
I've tried:
If s = 15, but you can't sub t in so how do you find out A?
If s = 0, t = 0, and you can't find A here.
What else do you know when t=0?
Original post by morgan8002
What else do you know when t=0?
initial velocity = 20
Do I differentiate the equation and put it equal to 20?
Original post by 2014_GCSE
initial velocity = 20
Do I differentiate the equation and put it equal to 20?
Do I differentiate the equation and put it equal to 20?
yeah
Original post by morgan8002
yeah
Thanks for helping me here
but urhh I'm still having trouble hahaso
S = A - Ae^-kt
V = kAe^kt
20 = kAe^k(0)
20 = kA
What have I done wrong...?
Original post by 2014_GCSE
Thanks for helping me here but urhh I'm still having trouble haha
so
S = A - Ae^-kt
V = kAe^kt
20 = kAe^k(0)
20 = kA
What have I done wrong...?
but urhh I'm still having trouble hahaso
S = A - Ae^-kt
V = kAe^kt
20 = kAe^k(0)
20 = kA
What have I done wrong...?
Sorry. I looked at the question quickly and didn't take into account the extra k when differentiating.
Sketch a graph of s against t. What happens as ? This is another common way of getting around the problem and removing the ks.
Original post by morgan8002
Sorry. I looked at the question quickly and didn't take into account the extra k when differentiating.
Sketch a graph of s against t. What happens as ? This is another common way of getting around the problem and removing the ks.
Sketch a graph of s against t. What happens as ? This is another common way of getting around the problem and removing the ks.
S gets bigger and moves towards the asymptote of A?
Original post by 2014_GCSE
S gets bigger and moves towards the asymptote of A?
Yes. The question says that the particle stops at s=15m, so this is the value of A.
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