How Do You Calculate Acceleration in Space Probe Motion Problems?

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To calculate acceleration in space probe motion problems, use the kinematic equation v^2 = vo^2 + 2a(x - xo), which does not require time. In this scenario, the probe's initial velocity is 2940 m/s, and its final velocity is 8820 m/s, with a displacement of 176.4 km converted to meters. After correcting unit discrepancies, the calculated acceleration is approximately 196 m/s². Understanding the conservation of energy can also aid in solving related parts of the problem. Accurate unit conversion and careful application of kinematic equations are crucial for obtaining the correct results.
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college phys 1111... neeed help!

Homework Statement



A space probe is desperately trying to escape the gravitational pull of a massive planet. After a final blast of its thrusters, the probe has a velocity of 2940 m/s away from the planet. Later, the probe is 176.4 km closer to the planet than it originally was, with a velocity of 8820 m/s toward the planet.

You are to find (assume constant acceleration):
a) the acceleration (magnitude and direction)
b) How far beyond its starting point the probe got.
c) How much time elapsed between initial and final positions.

I don’t understand how to approach this problem…. How do I find acceleration when I don’t know the time?
 
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You know that the probe's velocity was originally 2940 m/s, but that after falling 176.4 km towards the planet (and losing a corresponding amount of potential energy), its velocity became 8820 m/s. You can use the conservation of energy to calculate g, the acceleration due to gravity.

For b, you can use the conservation of energy again. c should be easy once you get the answer for a.
 


i am sorry i still don't understand... =[
 


i mean i know (Velocity final- Velocity initial)/time = acceleration but they don't tell the time and that is what is throwing me off...
 


efaizan said:
How do I find acceleration when I don’t know the time?

Look at the kinematic equations for motion with constant acceleration. Look for the one equation that does not contain time t.
 


v^2=vo^2+2a(x-xo) or
 


v^2-vo^2=2a(x-xo)
 


2940^2-8820^2=2a(x-176.4) and solve for a?
 


Yes. Um, you do know what the value of x is, right?
 
  • #10


ummm 0?
 
  • #11


Yup, exactly. But you used 176.4, which is in kilometers, whereas all your speeds are in m/s. You'll have to make the units agree.
 
  • #12


ok so 2940^2-8820^2=2a(0-109.6) i got a= 315459.85... that doesn't sound right does it?
 
  • #13


thank you guys btw for ur help!
 
  • #14


Well first convert x into meters, there are 1000 meters in 1km. Know you just plug and chug. And for some odd reason you have Vo as 8820 and V as 2940.
 
  • #15


ohhhh oops i think i converted the 176.4 km to miles... soooo i should flip flop the v's? and 176.4 km to meters would be 176400?
 
  • #16


ok so i did it again and i got 196 for a now... that sounds kinda right. right?
 
  • #17


Looks good.
 
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