The position of a particle moving along the x axis is given in centimeters by x=9.75+1.50t³, where t is in seconds. Calculate(a)
the average velocity during the time interval t=2.00 s to t=3.00 s(b) the instantaneous velocity at t=2.00 s(c)the instantaneous velocity at t=3.00 s(d)the instantaneous velocity at t=2.50 s(e)the instantaneous velocity when the particle is midway between its positions at t=2.00 s and t=3.00 s.(f) graph it
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Answer:7 cm/s
Explanation:
Given
Particle move along curve
As it reaches the (2,3) its y coordinate is increasing at 14 cm/s
Differentiating y w.r.t time
Now at (2,3)
The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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