Anyone that can pls help me with these 2 q's? (1 Viewer)

181jsmith · May 24, 2012

Below are the questions
thanks

1. A 6m long ladder is leaning up against a wall at a height of h and angle theta (like a right angled triangle). The ladder slips down the wall at a constant rate of 0.05m/s. Find the rate at which the angle is changing in degrees and minutes when the height is 2.5m.

2. The acceleration of a particle is given by $\ddot{x}= \frac{1}{1+x^2}$
where x is the displacement over time t seconds. If the particle starts at rest at the origin and always has positive velocity, find the equation of its velocity in terms of x

Drongoski · May 24, 2012

Q1 You draw the diagram

$h = 6sin \theta \implies \frac {dh}{dt} = -0.05 ms^{-1} = 6cos \theta \frac{d \theta}{dt}$

When h=2.5 sin(theta) = 2.5/6 = 5/12 ==> cos(theta) = sqrt(119)/12

$\therefore 0.05 = 6 \times \frac{\sqrt 119}{12} \times \frac{d \theta}{dt} \implies \frac{d\theta}{dt} = -\frac {0.05 \times 12}{6 \times \sqrt {119}} rads/sec = 0^ {\circ} 31' 31'' s^{-1}$$

wonder if answer is correct ????

Q2

$\ddot x = \frac{d}{dx}\frac {1}{2}v^2 = \frac {1}{1+x^2} \implies \frac {1}{2}v^2 = \int \frac {1}{1+x^2}dx = tan^{-1}x + C$

When t=0: v=0 x=0 ==> C = 0

$\therefore \frac {1}{2}v^2 = tan^{-1}x \implies v = + \sqrt{2tan^{-1}x}$

Carrotsticks · May 24, 2012

I am getting too slow and sloppy:

1. Drongoski beat me to it, with LaTeX!

2. Misread the question and derped.

Drongoski · May 24, 2012

carrot - you do the hard ones and I do the easier ones always. Besides I cheated by entering only 1 line first.

Carrotsticks · May 24, 2012

And yep your answer is correct. I got the exact same thing.

bleakarcher · May 24, 2012

181jsmith said:
Below are the questions
thanks

1. A 6m long ladder is leaning up against a wall at a height of h and angle theta (like a right angled triangle). The ladder slips down the wall at a constant rate of 0.05m/s. Find the rate at which the angle is changing in degrees and minutes when the height is 2.5m.

2. The acceleration of a particle is given by $\ddot{x}= \frac{1}{1+x^2}$
where x is the displacement over time t seconds. If the particle starts at rest at the origin and always has positive velocity, find the equation of its velocity in terms of x

If the ladder is leaning against the wall at an angle theta wouldnt that be the angle between the ladder and the wall...not the ladder and the floor?

Carrotsticks · May 24, 2012

bleakarcher said:
If the ladder is leaning against the wall at an angle theta wouldnt that be the angle between the ladder and the wall...not the ladder and the floor?

I interpreted it to be between the floor and the ladder, but realistically in an exam they would be more clear with it.

181jsmith · May 24, 2012

thanks heaps Drongoski and Carrot

Anyone that can pls help me with these 2 q's? (1 Viewer)

181jsmith

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Drongoski

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Carrotsticks

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Drongoski

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181jsmith

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