I need help with part b

[hollow sail]

I need help with part B

[plush fable]

I need help with part b

[stuck salmon]

Hi @hollow sail , What does L mean in this context? Speed of light?

[hollow sail]

This is just a math question nothing to do with physics

I have already found part a

v(t) = 4.9e^-t/2

[stuck salmon]

Alright, to determine the smallest possible value of L we need to find the maximum value of v(t):

lim x -> inf v(t)

[hollow sail]

So we need to find the max value of v(t)

I am allowed to use the calculator for this question from my teacher

[hollow sail]

How do I do this?

[stuck salmon]

I see that you have a small mistake in your velocity v(t). When integrating, we still get an integration constant:

v(t) = 19.6e^-t/2 + C

You can determine this by the boundary condition that the object starts falling at t=0 and therefore has no velocity yet:

v(t = 0) = 19.6e^-t/2 + C = 0
C = -19.6

Now you can calculate the limit. Consider what happens to e^-t/2 when t becomes very large

update: corrected the mistake

[hollow sail]

Oh I realized my mistake, let me fix the problem for part a

[hollow sail]

Ok so I have corrected by mistake

v(t)=19.6e^t/2 + C

[stuck salmon]

dont forget the minus in e^-t/2

[hollow sail]

ok

[stuck salmon]

now you can find C by the boundary condition: v(t=0) = 0. So C is -19.6 m/s

[hollow sail]

So that is that answer

[hollow sail]

Why is t=0 ?

[stuck salmon]

No, for the answer, you still need to calculate the limit of v(t) as t approaches infinity. This will give you the maximum velocity. So far, we have only done task a

[hollow sail]

oh

[hollow sail]

So how do I do this ? for task b

[stuck salmon]

In the problem, t is the time the object is falling. We release it at t = 0, at which point the object initially has no velocity since it first accelerates downwards. Therefore v(t=0) = 0

[hollow sail]

Oh that why

So how to solve part b

[stuck salmon]

We need to calc

$lim_{t \to \infty} v(t) = lim_{t \to \infty} 19.6e^{-\frac{t}{2}} - 19.6 $

wel..

[hollow sail]

well

$lim{t \to \infty} v(t) = lim{t \to \infty} 19.6e^{-\frac{t}{2}} - 19.6 $

[stuck salmon]

we need to calc

lim(t -> infinity) v(t)

lim(t -> infinity) 19.6e^-t/2 - 19.6

we can seperate it

lim(t -> infinity) 19.6e^-t/2 - lim(t -> infinity) 19.6

for t -> infinity we have in general

e^-t -> 0

therefore

lim(t -> infinity) 19.6e^-t/2 = 0

that means:

lim(t -> infinity) v(t) = 19.6 m/s

[hollow sail]

So what is the answer?

I have never used infinity before

[stuck salmon]

L = 19.6 m/s

[hollow sail]

Thank you

for helping me

IS there a another why to slove this not using infinity

?

[stuck salmon]

Oh okay, good to know. Then here's another possible solution:

We are looking for the maximum value of v(t) consists of two terms, where the first one has an e^-t/2 factor and the second one is a constant number. As t becomes larger and larger, the e^-t/2 factor becomes smaller and smaller, making the entire first term diminish until it eventually disappears. However, the second term remains constant throughout

[hollow sail]

Why is the second term remains constant?

[stuck salmon]

Since the second term is not dependent on t and always remains 19.6, regardless of how large or small t is

[hollow sail]

Ok so the answer is 19.6 m/s

[stuck salmon]

👍 yes

[hollow sail]

ok thank you for your help.

.close