An airplane awaiting departure is cleared to takeoff, its accelerates down a runway at 3.20m/ s2 for 32.8s until it finally lift
1 answer:
The final velocity before takeoff is 104.96 m / s.
<u>Explanation:</u>
The last velocity of a given object over some time defines the final velocity. The final velocity of the object is given by the product of acceleration and time and adding this product to the initial velocity.
To calculate the final velocity,
V = u + at
where v represents the final velocity,
u represents the initial velocity,
a represents the acceleration
t represents the time taken.
v = 104.96 m / s.
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Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
