Do I need to consider mass in the equations?

  • #1
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Homework Statement


a. At launch a rocket ship has a mass M. When it is launched from rest, it takes a time interval t1 to reach speed v1; at t = t2, its speed is v2.
What is the average acceleration of the rocket (i) from the start to t1 and (ii) between t1 and t2?

For now, I am taking question a as example, but these are the follow up questions:

(b. Assuming the acceleration is constant during each time interval (but not necessarily the same in both intervals), what distance does the rocket travel (i) from the start to t1 and (ii) between t1 and t2?

c. Use your symbolic results to calculate the average accelerations and distances for M = 2.25 * 106 kg, t1 = 8.00 s, v1 = 158 km/h, t2 =1.00 min, and v2 = 1580 km/h.)

Homework Equations


a= Δv/Δt
F = M * a?
s = vavg * Δt (in follow up questions)

The Attempt at a Solution


a) a= Δv/Δt
-> aavg = Δv/(2*Δt)
i) aavg = v1/(2*t1)
ii) aavg = (v2-v1)/(2*(t2-t1))

Is this correct or did I need to use mass M in the equations?
 

Answers and Replies

  • #2
ZapperZ
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1. No, you don't have to use the mass to find the average acceleration. The force is already acting on the mass to produce such acceleration, so mass is already taken into account, resulting in those change in velocities.

2. Where did you get the "2" in aavg = Δv/(2*Δt) ? Why is it there?

Please note that aavg is defined as the change in velocity over that time period, i.e. Δv/Δt. It appears that you're taking a "statistical" definition of "average" here, which may account for the appearance of the "2". This is NOT how it is defined in physics.

Zz.
 
  • #3
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Where did you get the "2" in aavg = Δv/(2*Δt) ? Why is it there?
Oh, yes, I see I made a mistake in thinking that dividing by 2 got me the average velocity.

Thank you for the answer!
 
  • #4
jbriggs444
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It appears that you're taking a "statistical" definition of "average" here, which may account for the appearance of the "2".
The notions of average are the same in both disciplines. The distinction is between averaging over a uniformly weighted discrete distribution of two elements (sum the element values and multiply by the uniform weight of 0.5) and a time-weighted continuous distribution (e.g. integrate the acceleration over time from start to finish and divide by elapsed time from start to finish).

Of course, if you integrate acceleration over time from start to finish, you get delta v between start and finish. So you get a formula that looks different, but ultimately it is the same concept.
 
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