Mechanics As level M1find the acceleration of the flare

Struggling like mad with this CIE As M1 question: A flare is launched from a hot air balloon and moves in a vertical line. At time t seconds , the height of the flare is x metres, where x=1664-40t-2560÷t for t> (or equal to ) 5s.
The flare is launched when t=5.
a) Find the height and the velocity of the flare immediately after it is launched.
b) Find the acceleration of the flare immediately after it is launched, when its velocity is zero, and when t=25
c) Find the terminal speed of the flare.
d) Find when the flare reaches the ground.
e) Sketch the (t,v) graph and the (t,x) graph for the motion of the flare.

I got a),b) I'm not sure what it means to have an acceleration when v=o. However c) I only got by imagining t goes to infinity. right??d) I just can't get. I know the initial speed when the flare leaves & I know terminal speed ( answer to c) but since a is decreasing until it reaches zero, I can't work out the time it takes to go from initial speed to terminal speed; once I know that I'm home & dry, since we are given h (answer to a)i. I would really appreciate a step by step working so I can really understand what is going on here. Thanks :smile:

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Reply 1

qwert7890

Original post by Yerffoeg

the height of the flare is 1664-40t-(2560÷t)??
Because at t=5, x = 952 which is weird because if it is launched at t=5, then x = 0....

Reply 2

davros

Original post by Yerffoeg

Can you upload a picture of the question, because your equation for x doesn't look right as mentioned above!

Reply 3

mqb2766

Original post by qwert7890

the height of the flare is 1664-40t-(2560÷t)??
Because at t=5, x = 952 which is weird because if it is launched at t=5, then x = 0....

it is from a hot air balloon, so x(5) can be nonzero.

Reply 4

mqb2766

Original post by Yerffoeg

b) when you throw a ball upwards, gravity/acceleration is -9.8, even when v=0 at its highest point. Acceleration is the rate of change of velocity. The velocity goes from positive to negative at its highest value, so acceleration musr be non zero.
c) terminal speed is the limiting value, so t tending to infinity.
d) when is x=0, its that easy? You seem to be, incorrectly, describing a suvat-type solution?

All this is assuming the flare goes through the hot air balloon at the start. But thats maths for you.

(edited 2 years ago)

Reply 5

Yerffoeg

sorry i don't know how to send image. Thanks. I checked the equatiion and it's right.

Reply 6

Yerffoeg

Original post by davros

Can you upload a picture of the question, because your equation for x doesn't look right as mentioned above!

I'm sorry but I don't know how to do that. i tried dragging photo onto this space but it didn't work. I checked the equation of x and it is right.

Reply 7

Yerffoeg

Original post by Yerffoeg

I'm sorry but I don't know how to do that. i tried dragging photo onto this space but it didn't work. I checked the equation of x and it is right.

Thanks for your replies. I idiotically got it into my head that the flare was initially thrown downwards...now it all makes sense. thanks

Reply 8

Yerffoeg

Original post by Yerffoeg

Thanks for your replies. I idiotically got it into my head that the flare was initially thrown downwards...now it all makes sense. thanks

Hi again..sorry but the correct answer to b)I, the acceleration of the flare immediately after it is launched, is given as negative 41ms^-s.... This just threw me, coz I had understood the flare had been fired up vertically ...please can someone enlighten me as to why this acceleration here is negative....that's why I later thought the flare was fired downwards...am I right? Thanks

Reply 9

mqb2766

Original post by Yerffoeg

If upwards is positive, as it is in this question, what is the acceleration due to gravity?

(edited 2 years ago)

Reply 10

Yerffoeg

Original post by mqb2766

If upwards is positive, as it is in this question, what is the acceleration due to gravity?

ok the accel due to gravity is -g, what I don't get is why it is bigger than g..

Reply 11

mqb2766

Original post by Yerffoeg

ok the accel due to gravity is -g, what I don't get is why it is bigger than g..

So youre happy with the negative acceleration decreasing the velocity? The velocity is positive initially, decreases to 0 when it is at the max height, then the velocity becomes negative as it plunges back to earth. The trajectory is very roughly quadratic with negative curvature/acceleration.

Why what is bigger than g?

(edited 2 years ago)

Reply 12

Yerffoeg

Original post by mqb2766

So youre happy with the negative acceleration decreasing the velocity? The velocity is positive initially, decreases to 0 when it is at the max height, then the velocity becomes negative as it plunges back to earth. The trajectory is roughly quadratic with negative curvature/acceleration.

Why what is bigger than g?

The acceleration of the flare immediately after it is launched. The correct answer is a negative number and much bigger than g

Reply 13

Yerffoeg

Original post by Yerffoeg

The acceleration of the flare immediately after it is launched. The correct answer is a negative number and much bigger than g

It is the first part of the answer to b)

Reply 14

mqb2766

Original post by Yerffoeg

The acceleration of the flare immediately after it is launched. The correct answer is a negative number and much bigger than g

Two things:
1) The height expression is a math model with an easy closed form formula. Its not necessarily a perfect model, especially when you differentiate twice.
2) However, more physically realistic, another force is acting on the flare which is producing a large deceleration initially (when the velocity is large) as well as gravity. What might it be?

(edited 2 years ago)

Reply 15

Yerffoeg

Original post by mqb2766

Please give me a hint...I don't know!!!

Reply 16

mqb2766

Original post by Yerffoeg

Please give me a hint...I don't know!!!

It also causes the terminal velocity behaviour when the flare is descending. This is when the acceleration is zero (rather than -g). Its gcse physics.

(edited 2 years ago)

Reply 17

Yerffoeg

Original post by Yerffoeg

Please give me a hint...I don't know!!!

..I'll try ...before it is released it is attached to the balloon..sorry really..I don't know ..do I need 6to cite Newton's third?

Reply 18

mqb2766

Original post by Yerffoeg

..I'll try ...before it is released it is attached to the balloon..sorry really..I don't know ..do I need 6to cite Newton's third?

Nothing to do with the balloon. To be honest if you google
terminal velocity
The first link (wiki) tells you straight away. As this a question part, I'll leave you to have a read.

However, its worth noting that I was being way too critical of the math model earlier. The height is maximum at t=8, so the velocity is zero at that time point. At this point, the acceleration is -10 so -g. So the additional force is zero when the velocity is zero as it should be. So actually the math model of the height is pretty good.

(edited 2 years ago)

Reply 19

Yerffoeg

thanks...Yes, I googled terminal velocity as you suggested.The answer I guess you wish me to say is air resistance. I'm still confused though.. When v=o, g was acting down on it. But here after 5 seconds is the acceleration acting downwards on it due to the air resistance while it is on its last phase of going up, with a.r acting in the opposite direction downwards opposing the upwards motion? Thanks