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MArishka [77]
2 years ago
11

an airplane is moving at a speed of 75 m/s. as it. lands on a. runwsys. if. the runway is. 500m long, what is the acceleration o

f the plane as it slows to a stop?
Physics
1 answer:
bezimeni [28]2 years ago
4 0

Answer: 0.075

explanation:

you can use one of the equations of motion.

2as =v^2 - u^2

we shall make acceleration the subject as that's what we need to find out

a = (v^2 - u^2) / (2 × s)

now substitute the given values:

final speed would be zero as the airplane stops.

a = (75 - 0) / ( 2 × 500)

a = 75 ÷ 1000

a = 0.075 meters per second squared.

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Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

v=u+at

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec²  [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2

\Rightarrow 16\times t^2=576

\Rightarrow t^2=\frac{576}{16}

\Rightarrow t^2=36

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

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