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3 years ago

An airplane traveling 245 m/s east experiences turbulence, so the pilot slows down to 230 m/s. It takes the pilot 7

Physics

2 answers:

makvit [3.9K]3 years ago

8 0

Answer:a

Explanation:

USPshnik [31]3 years ago

5 0

Answer:

The acceleration of the plane is -2.14 m/s² ⇒ 2nd answer

Explanation:

Acceleration is the rate of change of velocity

The unit for acceleration is meters per second square

The rule of the acceleration is: → $a=\frac{v-u}{t}$

where a is the acceleration, u is the initial velocity, v is the final velocity

and t is the time

An airplane traveling 245 m/s east experiences turbulence so the pilot

slows down to 230 m/s

It takes the pilot 7 seconds to slow down to this speed

We need to find the acceleration of the plane

Lets use the rule above

→ u = 245 m/s , v = 230 m/s , t = 7 seconds

→ $a=\frac{230-245}{7}$

→ $a=\frac{-15}{7}=-2.14$

→ The acceleration is -2.14 m/s²

The acceleration of the plane is -2.14 m/s²

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A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

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We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$