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Fofino [41]
4 years ago
14

Your car is stalled in the middle of a large patch of ice (assumed to be frictionless). You have a friend that has thrown a rope

to you and you attach it to the car.Your friend then applies a continuous horizontal force of 550 N to you and the car. If you and the car have a total mass of 1430 kg, how long will it take for you to reach
Physics
1 answer:
Gnoma [55]4 years ago
7 0

Answer:

The time depends on the distance that they have to travel

x(t) = \frac{0.3846t^{2} }{2}

Explanation:

The only horizontal force exerts over the car and you, it is the force that your friend is applied

Newton's Second Law of Motion defines the relationship between acceleration, force, and mass, thus

\sum{F} = ma

        550 = 1430a

            a = 0.3846 m/s2

             

The car and you have a motion under constant acceleration, then theirs position to a time-based is:

x(t) = x_{0} + v_{0}t +\frac{at^{2} }{2}

By the initial conditions

x(t) = \frac{at^{2} }{2}

x(t) = \frac{0.3846t^{2} }{2}

The time depends on the distance that they have to travel  

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Answer:

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Explanation:

Given that,

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Velocity of boat = 5.60 m/s due north

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We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

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The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

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For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

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For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

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v_{sb_{x}=2.-1.82

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Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

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