Simple constant acceleration/equation of motion question

  • #1

Homework Statement


A 747 Jumbo Jet must reach a speed of 290km/h by the end of the runway to lift off. The 15L-33R runway at Toronto International Airport is 2,770 meters long. The main Toronto Island Airport runway 08-26 is 1216 meters long. Assuming the jet has a constant acceleration, for each runway:
a) Determine the time to reach the end of the runway
b) Determine the required acceleration.
c) Do you think 747’s will ever take off and land on the island? Explain in 3 or 4 lines

Homework Equations


t = d/s
v = u + at
v^2 = u^2 + 2ax
where u = initial velocity, x = displacement

The Attempt at a Solution


a) 290 km/h = 80.5 m/s

Runway A)
t = 2770/80.5
= 34.4 s

Runway B)
t = 1216/80.5
= 15.1 s

b) Can I assume initial velocity to be 0 and then just use the formulas? Can I also just use the constant acceleration formulas to calculate time for part a)? Because when I do try that and input initial velocity as 0, then the values I get for time are exactly two times the values I got in part a) using t = d/s. Why is this the case?

Thanks a lot.
 

Answers and Replies

  • #2
Doc Al
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Because when I do try that and input initial velocity as 0, then the values I get for time are exactly two times the values I got in part a) using t = d/s. Why is this the case?
Because when you use t = d/s you are assuming a constant speed. But that's not the case. (Hint: What's the average speed of the jet as it accelerates?)
 
  • #3
Because when you use t = d/s you are assuming a constant speed. But that's not the case. (Hint: What's the average speed of the jet as it accelerates?)
Ah, I see. In that case,

Runway A)
x = 2770 m
u = 0 m/s
v = 290 km/h = 80.5 m/s

v^2 = u^2 + 2as

80.5^2 = 0 + 2a(2770)
a = 1.17 m/s^2

Calculating t:
v = u + at
80.5 = 0 + (1.17)(t)
t = 68.8 s

this is my revised solution. Would this method suffice?
 
  • #4
Doc Al
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Would this method suffice?
Sure.

But it might even be easier to use d = vavet to find the time. Then v = at to find the acceleration. But your way is fine too!
 
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  • #5
Sure.

But it might even be easier to use d = vavet to find the time. Then v = at to find the acceleration. But your way is fine too!

Actually, just took another look at this. Would you mind explaining how you'd use your method to do this? Particularly d = vavet. I feel like that's the same as 2770 meters/250 km/h, no? As that is what I did initially (albeit using scalar quantities, t = d/s) but that was under the incorrect assumption that speed was constant.

Thanks a lot!
 
  • #6
PeroK
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Actually, just took another look at this. Would you mind explaining how you'd use your method to do this? Particularly d = vavet. I feel like that's the same as 2770 meters/250 km/h, no? As that is what I did initially (albeit using scalar quantities, t = d/s) but that was under the incorrect assumption that speed was constant.

Thanks a lot!

First, the time for a journey is distance/average speed. So, if you know any two of distance, time and average speed, you can simply calculated the one you don't know.

Second, for constant acceleration, the average speed is half way between the initial speed and the final speed. So, if you start at 10m/s and constantly accelerate to 30m/s, then your average speed during this time is 20m/s.

In particular, if you accelerate from rest, then the average speed is half the final speed.

Note: you used the final speed in your calculation (which is clearly not right) and @Doc Al used the average speed.
 
  • #7
Doc Al
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Would you mind explaining how you'd use your method to do this?
I think @PeroK did a great job explaining how average velocity is used. Let us know if anything is unclear.
 

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