# Jet accelerating in problem with Constant Acceleration

1. May 5, 2014

### Caolan

1. The problem statement, all variables and given/known data
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engine up to full throttle. After travelling 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

2. Relevant equations
t1 = x1 - x0 / v1 - v0
a = v1 - v0 / t1

3. 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

2. May 5, 2014

### Staff: Mentor

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.)

3. May 5, 2014

### 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?

4. May 5, 2014

### 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!

5. May 5, 2014

### 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.

6. May 5, 2014

### Staff: Mentor

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.

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

That's fine.

7. May 5, 2014

### Staff: Mentor

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

8. May 5, 2014

### 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?

9. May 5, 2014

### goraemon

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.

10. May 5, 2014

### Caolan

awesome thank you!!!

11. May 5, 2014

### Staff: Mentor

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 = vave*Δt.

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