# physics

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?

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1. 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?

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2. 10.48m

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3. 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 :)

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4. 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)

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