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harkovskaia [24]
3 years ago
6

A cyclist maintains a constant velocity of

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

Hi there!

We can use the equation:

d = x₀ + vt, where:

x₀ = initial distance from the reference point

v = velocity (m/s)

t = time (sec)

Plug in the given values:

d = 248 + 5(49)

d = 493m

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Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

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D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

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R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

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-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

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D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

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