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2 years ago

11

A plane has a takeoff speed of 88.3 m/s and can accelerate at a rate of 3m/s to reach that speed. How long does the plane take t

o take-off.

1 answer:

Mrrafil [7]2 years ago

3 0

v = $\frac{d}{t}$

and

a = $\frac{v}{t}$

We have acceleration and velocity so:

3 = $\frac{v}{t}$

88.3 = $\frac{d}{t}$

In the acceleration equation we can isolate for v and then plug it back into the other equation to solve...

So...

$3t = v$

$88.3 = 3t$

Divide by three and

t = 29.4 s

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The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

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Where:

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$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

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<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

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