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9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Skull is moving relative to himself. If Red Skull is running at 3.50 m/s at an angle of 45.0° toward the left front of a truck that is moving at 12.5 m/s N, and Captain America is running at 4.00 m/s at an angle of 45.0° to the front left of a second truck moving South at 15.0 m/s, find the relative velocity for Captain America to hit Red Skull. a​

Respuesta :

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately 33.23 m/s in a direction from the North of

approximately 9.18°.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

  • v₁ = -([tex]\frac{\sqrt{2} }{2}[/tex] × 3.5)·i + ([tex]\frac{\sqrt{2} }{2}[/tex] × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

  • v₂ =  ([tex]\frac{\sqrt{2} }{2}[/tex] × 4.0)·i - ([tex]\frac{\sqrt{2} }{2}[/tex] × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-([tex]\frac{\sqrt{2} }{2}[/tex] × 3.5) - ([tex]\frac{\sqrt{2} }{2}[/tex] × 4.0))·i + (([tex]\frac{\sqrt{2} }{2}[/tex] × 3.5 + 12.5) + ([tex]\frac{\sqrt{2} }{2}[/tex] × 4.0 + 15))·j

  • v₁₂ = [tex]-\frac{ 15 \cdot \sqrt{2} }{4}[/tex]·i + [tex]\frac{ 110 + 15 \cdot \sqrt{2} }{4}[/tex]·j

Red Skull's velocity relative to Captain America,  v₁₂ = [tex]-\frac{ 15 \cdot \sqrt{2} }{4}[/tex]·i + [tex]\frac{ 110 + 15 \cdot \sqrt{2} }{4}[/tex]·j

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

  • Red Skull appears to be moving West at 5.3 m/s and North at 32.8 m/s

  • The direction is [tex]arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}[/tex]

Therefore;

  • Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

  • [tex]|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23[/tex]·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ 33.23 m/s

Learn more here:

https://brainly.com/question/24430414

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