Most hummingbirds can fly with speeds of nearly 50.0 km/h. Suppose a hummingbird flying with a velocity of 50.0 km/h in the forward direction accelerates uniformly at 9.20 m/s^2 in the backward direction until it comes to a hovering stop. What is the hummingbird's displacement?
vi = 50 km/h = 13.9 m/s
vf = 0
a = -9.20 m/s^2
d = ?
vf^2 = vi^2 + 2ax
0=13.9^2 + 2(-9.20)x
0 =193 - 18.4x
18.4x = 193
divide both sides by 18.4
x = 10.5
displacement = 10.5 m
notice: (all measurements are rounded to the smallest number of significant figures)
Initial velocity, u = 50 km/h = 13.89 m/s
Final velocity, v = 0 km/h = 0 m/s
Acceleration, a = -9.2 m/s
Displacement, S
2aS = v²-u²
S=(v²-u²)/(2a)
Can you do the rest?
10.48m
Well, let's put our humor goggles on and figure this out. A hummingbird who's a speed demon, flying all fast and furious at 50.0 km/h, suddenly decides to hit the brakes with a backward acceleration of 9.20 m/s^2. It's like the bird got a sudden urge for a reverse gear!
Now, to calculate the displacement, we need to know the time it takes for Mr. Hummingbird to come to a stop. Unfortunately, that vital info is missing from the equation. So, let's assume that this speed demon-to-hovering expert takes one minute to bring itself to a standstill. Crazy, I know!
In one minute, our hummingbird would have traveled 50.0 km/h, which is about 833.33 meters. Since it starts and stops at the same position, the displacement is zero. Yep, absolutely nada. In physics terms, this means the little birdie goes nowhere in the end.
Just remember, even though the hummingbird didn't go anywhere, it still swerved some serious physics moves!
To find the hummingbird's displacement, we can use the equations of motion for uniformly accelerated motion.
First, let's convert the hummingbird's velocity from km/h to m/s. We know that 1 km/h is equal to 1000 m/3600 s, so:
Velocity = 50.0 km/h * (1000 m/3600 s) = 13.89 m/s
Now, we have the initial velocity (u) as 13.89 m/s, acceleration (a) as -9.20 m/s^2 (negative because it's in the opposite direction), and we need to find the displacement (s).
The equation we can use to find displacement is:
s = (v^2 - u^2) / (2a)
where v is the final velocity, and u is the initial velocity.
In this case, the hummingbird comes to a stop, so the final velocity (v) is 0 m/s. Plugging in the values, we have:
s = (0^2 - 13.89^2) / (2 * -9.20)
s = (-193.39) / (-18.40)
s = 10.51 meters
Therefore, the hummingbird's displacement is 10.51 meters.
we use the 4th kinematic equation :vf'2
=vi'2=2ax... but first we have to rearrange it :x+vf'2-vi /2a
+50'2 -0 /2(9.20)
+2500/18.4
+136km
hop this is right :)