Assistance needed with vector velocity problem, please

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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!