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lawyer [7]
3 years ago
13

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 Sku

ll 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​
Physics
1 answer:
Sedaia [141]3 years ago
6 0

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

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

approximately <u>9.18°</u>.

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 = -(\frac{\sqrt{2} }{2} × 3.5)·i + (\frac{\sqrt{2} }{2} × 3.5 + 12.5)·j

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

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

Relative velocity, v₁₂ = v₁ - v₂

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

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

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

  • v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

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

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

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;

  • |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·

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

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

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We can solve this problem by using the law of conservation of energy.

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W_f = -F_f d is the work done by friction (which is negative since the force of friction is opposite to the motion), where

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The speed of the ball is maximum at the instant when all the elastic potential energy stored in the spring has been released: in fact, after that moment, the spring does no longer release any more energy, so the kinetic energy of the ball from that moment will start to decrease, due to the effect of the work done by friction.

The elastic potential energy of the spring is

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(c) 1.78 m/s

The maximum speed is the speed of the ball at the moment when the kinetic energy is maximum, i.e. when all the elastic potential energy has been released.

As we calculated in part (a), the total energy released by the spring is

E = 0.0102 J

The work done by friction here is just the work done to cover the distance of

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Therefore

W_f = -(0.0323)(0.0508)=-1.64\cdot 10^{-3} J

So, the kinetic energy of the ball at the point of maximum speed is

K_f = E+W_f = 0.0102 - 1.64\cdot 10^{-3}=0.00856 J

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