Jet landing speed calculation: getting 333 km/hr instead of 666 km/hr

[pkossak]

I feel like I'm losing my mind asking such an easy question, but for some reason I can't think of why the answer is double of what I'm calculating every time for this question.

After landing, a jet airplane comes to rest uniformly
(the acceleration is constant) in 11.5 seconds. The aircraft
rolls 1063.75 m. What was the landing speed? (in km/hr)

I keep getting 333 km/hr, but the answer is listed as 666 km/hr.
thanks

 

[mukundpa]

s = average speed x time
= (vinitial + vfinal)/2 x time

 

[AKG]

What equation are you using? You should be able to derive the equation that relates initial velocity, final velocity, time elapsed, and displacement. From this equation, you should be able to find the initial velocity.

 

[pkossak]

i was using x = xo + Vav*t

 

[pkossak]

And I don't know of an equation I can use for all that without having to use acceleration.

for the Vav = .5(v +vo), I don't see how I could I plug in 333 and come out with 666, as 333 would be v, and 0 would be vo.

I can't use x = xo + vo*t + .5*a*t^2 because I don't know acceleration

why am i having all this trouble

 

[Astronuc]

Consider that the acceleration, or deceleration as the case may be, is uniform. This gives the total distance traveled as a function of acceleration and time. Hint: $\ddot{x} = -a$.

Then how about $v_f^2 = v_i^2 + 2 a x$ make sure you get the magnitude of a correct.

 

[AKG]

If this is a college level course, you should be able to derive the right equation.

$$\Delta d = \int _{0} ^{T} v(t)\, dt$$

$$\Delta d = \int _{0} ^{T} v_i + at\, dt$$

$$\Delta d = v_iT + 0.5aT^2$$

$$\Delta d = v_iT + 0.5(v_f - v_i)T$$

$$\frac{\Delta d}{T} = \frac{v_f + v_i}{2}$$

This says that the average velocity starting from time 0 to any time T is just the average of the final and initial velocities (given constant acceleration). If this is a high school course, then you should be given this (and the other 4) kinematics equations.