In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. An
swer in m/s and the number only to the
nearest 100th (ie. The second position after the decimal point.). Remind yourself of the definition of velocity.
nearest 100th (ie. The second position after the decimal point.). Remind yourself of the definition of velocity.
1 answer:
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Answer:
a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s
Explanation:
a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration
Fe + Fm = m a
a = (Fe + Fm) / m
the electric force is
Fe = q E k ^
Fe = 4.25 10-4 60 k ^
Fe = 2.55 10-2 k ^
the magnetic force is
Fm = q v x B
Fm = 4.25 10⁻⁴
fm = 4.25 10⁻⁴ (-j ^ 30 4)
fm 0 = ^ -5,10 10⁻² j
We look for every component of acceleration
X axis
aₓ = 0
there is no force
Axis y
ay = -5.10 10²/5 10⁻⁵ j ^
ay = -1.02 107 j ^ / s2
z axis
az = 2.55 10⁻² / 5 10⁻⁵ k ^
az = 5.1 10² k ^ m / s²
Having the acceleration in each axis we can encocoar the position using kinematics
X axis
the initial velocity is vo = 30 m / s and an initial position xo = 0
x = vo t + ½ aₓ t₂2
x = 30 0.02 + 0
x = 0.6m
Axis y
acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m
y = I + go t + ½ ay t²
y = 1 + 0 + ½ (-1.02 10⁷) 0.02²
y = 1 - 2.04 10³
y = -2039 m j ^
z axis
acceleration is aza = 5.1 10² m / s², the position and initial speed are zero
z = zo + v₀ t + ½ az t²
z = 0 + 0 + ½ 5.1 10² 0.02²
z = 1.02 10⁻¹ m k ^
therefore the position of the bodies is
r = (0.6 i- 2039 j ^ + 0.102 k⁾ m
b) x axis
since there is no acceleration the speed remains constant
vₓ = 30.0 m / s
Axis y
let's use the equation v = v₀ + t
= 0 + -1.02 10⁷ 0.02
v_{y} = 2.04 10⁵ m / s
z axis
v_{z} = vo + az t
v_{z} = 0 + 5.1 10² 0.02
v_{z} = 1.02 10⁻¹m / s