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astra-53 [7]
4 years ago
9

A person is walking on a treadmill. The treadmill is moving backwards at a speed of 3.0 m/s. The person is moving forward relati

ve to the treadmill at 2.5 m/s. What is the person’s velocity relative to the ground?
Physics
1 answer:
Serga [27]4 years ago
4 0

Answer:

V = - 0.5[m/s]

Explanation:

To solve this problem we have to analyze it by means of relative speeds, we have a treadmill that moves at a speed of -3 [m/s], (the negative sign means that it moves in a sense opposite to that of the person). The person moves at a speed of 2.5 [m/s]

Therefore we have from the perspective of a person placed outside

- 3 + 2.5 = - 0.5 [m/s]

Note: This would physically look like the person retreating as he or she walks forward, should increase his or her speed to 3 [m/s] to balance the speed of the band and not get out of it.

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Answer:

a) -2.34 m

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Explanation:

part a

Given:

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t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

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part b

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part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

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0.14 m/s = - A*w*sin (w*t)  .....Eq2

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Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

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