Assistance needed with vector velocity problem, please

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SUMMARY

An asteroid traveling at 15 km/s was modified with a rocket engine, resulting in a new velocity of 19 km/s at an angle of 28° after 10 minutes. The average acceleration was calculated using the change in velocity, yielding (3.0i + 15j) m/s². The initial approach utilized the cosine law for magnitude but was corrected to use the coordinate method for direction, as suggested by the professor. This method is more efficient for vector calculations.

PREREQUISITES
  • Understanding of vector components and acceleration
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of kinematic equations for motion
  • Ability to convert units, particularly from km/s to m/s²
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  • Study the coordinate method for vector addition and subtraction
  • Learn about kinematic equations and their applications in physics
  • Explore the use of trigonometric identities in vector calculations
  • Practice problems involving acceleration and direction changes in motion
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Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators looking for teaching strategies in vector calculations.

SelHype
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An asteroid is discovered heading straight toward Earth at 15 km/s. An international team manages to attach a giant rocket engine to the asteroid. The rocket fires for 10 min, after which the asteroid is moving at 28[tex]\circ[/tex] to its original path at a speed of 19 km/s.

Find its average acceleration (ax, ay) in m/s2.

I first began by using the equation a2 = b2 + c2 -2bc(cos[tex]\alpha[/tex]) where b is 15 km/s and c is 19 km/s.

a2 = 225 + 361 - 570(cos28[tex]\circ[/tex])
a2 = 82.7 km/s
9.1 km/s m= [tex]\Delta[/tex]v

a= 9.1 / 600 = .0152 km/s2 = 15.2 m/s2 The answer is [tex]r\hat{}[/tex] = (3.0[tex]i\hat{}[/tex] + 15 [tex]j\hat{}[/tex]) m/s2.

I am unsure as to whether or not I have done this correctly because I do not know where to go from here. My professor gave use this hint for this problem:

The asteroid is initially going in the +x direction! From the given initial and final
velocities, find [tex]\Delta[/tex]vx and [tex]\Delta[/tex] vy. Use ax = [tex]\Delta[/tex]vx/[tex]\Delta[/tex]t and ay = [tex]\Delta[/tex]vy/[tex]\Delta[/tex]t
 
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SelHype said:
An asteroid is discovered heading straight toward Earth at 15 km/s. An international team manages to attach a giant rocket engine to the asteroid. The rocket fires for 10 min, after which the asteroid is moving at 28[tex]\circ[/tex] to its original path at a speed of 19 km/s.

Find its average acceleration (ax, ay) in m/s2.

I first began by using the equation a2 = b2 + c2 -2bc(cos[tex]\alpha[/tex]) where b is 15 km/s and c is 19 km/s.

a2 = 225 + 361 - 570(cos28[tex]\circ[/tex])
a2 = 82.7 km/s
9.1 km/s m= [tex]\Delta[/tex]v

a= 9.1 / 600 = .0152 km/s2 = 15.2 m/s2

The answer is [tex]r\hat{}[/tex] = (3.0[tex]i\hat{}[/tex] + 15 [tex]j\hat{}[/tex]) m/s2.

Hi SelHype! Welcome to PF! :smile:

The question asks for (ax, ay).

Your cosine formula only gave you the magnitude, |a| (which was correct :wink:) …

but you won't get the direction without using the sine formula also, which is far too long-winded a method.

There are two ways of dealing with vectors … the good old trigonometry way that the ancient Greeks would have used, and the coordinate method.​

You've used the slow ancient Greek way.

Your professor wants you to use the quicker coordinate way. :wink:

Do what your professor suggested … :smile:
 


tiny-tim said:
Hi SelHype! Welcome to PF! :smile:

The question asks for (ax, ay).

Your cosine formula only gave you the magnitude, |a| (which was correct :wink:) …

but you won't get the direction without using the sine formula also, which is far too long-winded a method.

There are two ways of dealing with vectors … the good old trigonometry way that the ancient Greeks would have used, and the coordinate method.​

You've used the slow ancient Greek way.

Your professor wants you to use the quicker coordinate way. :wink:

Do what your professor suggested … :smile:

Thank you for the welcome!

I should have known I was doing it the long way, haha. I am VERY bad for going the more complicated routes because...well they seem easier...Yeah I'm odd.

But thank you for the help! I finally got it after I looked at it for bout another hour, haha.

Anyways, thanks again!
 

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