Jet accelerating in problem with Constant Acceleration

[Caolan]

Homework Statement

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engine up to full throttle. After traveling 4km, the jet is moving 400 m/s. What is the jet's acceleration?

known:
x0 = 0m, t0 = 0s, v0 = 300 m/s
x1 = 4000m, t1 = ?, v1 = 400 m/s

Homework Equations

t1 = x1 - x0 / v1 - v0
a = v1 - v0 / t1

The Attempt at a Solution

t1 = x1 - x0 / v1 - v0 = 4000m / 100 m/s = 40s
a = v1 - v0 / t1 = 100m/s / 40s = 2.5 m/s/s

The book says the acceleration is 8.8m/s/s. I have worked this problem several ways and cannot seem to come up with the solution

 

[Doc Al]

t1 = x1 - x0 / v1 - v0 = 4000m / 100 m/s = 40s

This reasoning is not correct. To find the time it took to cover those 4 km you need the average speed during that interval. Which is what? (For some reason, you used the change in speed.)

 

[paisiello2]

In solving for t1 you incorrectly assumed that the velocity was a constant 100 m/s.

Do you another kinematic equation for distance under a constant acceleration?

 

[Caolan]

I assumed it was 100m/s because it starts off at 300 m/s and ends up at 400 m/s after 4000m and so the difference is 100 m/s. the other ones I have is the standard one:

x1 = x0 + v0(t1 - t0) + 0.5 * a(t1 - t0)^2. Substituting in the values I come up with:

4000 = 300(40) + 0.5 * a(40)^ = 4000 = 12000 + 0.5 * a * 1600
-8000 = 800*a
10 m/s/s = a.

Or I could use this one:
v1^2 = v0^2 + 2a * (x1 - x0)
400 ^ 2 = 300 ^ 2 + 2a * 4000)
160000 - 90000 = 8000a
70000 = 8000a

...

Okay I swear I did this several times... grrrr.

Thanks!

 

[Caolan]

So my question now is how do I know which of the three kinematic equations to use? All three provide through algebraic methods the ability to find the acceleration? But I have gotten different answers from all three. The ciorrect answer being in the third one.

 

[Doc Al]

I assumed it was 100m/s because it starts off at 300 m/s and ends up at 400 m/s after 4000m and so the difference is 100 m/s.

That's certainly the difference in speed, but that's not what you need to calculate the time. To calculate the time, you can use distance = ave speed * time, but first you need the average speed.

the other ones I have is the standard one:

x1 = x0 + v0(t1 - t0) + 0.5 * a(t1 - t0)^2. Substituting in the values I come up with:

4000 = 300(40) + 0.5 * a(40)^ = 4000 = 12000 + 0.5 * a * 1600
-8000 = 800*a
10 m/s/s = a.

This is wrong because you are assuming a time of 40 s, which is wrong.

Or I could use this one:
v1^2 = v0^2 + 2a * (x1 - x0)
400 ^ 2 = 300 ^ 2 + 2a * 4000)
160000 - 90000 = 8000a
70000 = 8000a

That's fine.

 

[Doc Al]

So my question now is how do I know which of the three kinematic equations to use? All three provide through algebraic methods the ability to find the acceleration? But I have gotten different answers from all three. The ciorrect answer being in the third one.

Your only mistake was using this equation, which is not valid:

t1 = x1 - x0 / v1 - v0

 

[Caolan]

I recalculated it using the average and yes I was able to get 8.8 m/s^2 again. So is it safe to say that whenever you're given two velocities and asked to find something between that you should find the average and calc based on that?

 

[goraemon]

I recalculated it using the average and yes I was able to get 8.8 m/s^2 again. So is it safe to say that whenever you're given two velocities and asked to find something between that you should find the average and calc based on that?

Technically, these kinematics equations only work when you know the acceleration is constant. But for practical purposes, virtually all kinematics questions assume constant acceleration, so yes.

 

[Caolan]

awesome thank you!

 

[Doc Al]

So is it safe to say that whenever you're given two velocities and asked to find something between that you should find the average and calc based on that?

Well, it depends on what you need to calculate and how you plan to do it.

If you want to use a kinematic formula that involves the time, then you'll need to calculate the time. For that you can use Δx = v_ave*Δt.

A useful list of kinematic formulas can be found here: Basic Equations of 1-D Kinematics