A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the least constant acceleration needed for takeoff form a 1.80 km runway?
I couldn't figure out which equation to use
I think there are a couple of equations involved here.
You're given s=1.8km and v_final =360km/hr
You'll need to look for equations that relate s,v,a and t
You might use s=(1/2)at^2 and v=at
You'll need to do some manipulating and substituting. You might also want to convert the km/hr to m/sec.
I'll be glad to check your work.
You seem to have figured most of this out already. However, I would recommend using the equation:
v^2 = 2as.
This brings you to:
10000 = 2*1800*a
so...
a=10000/3600 = 2.78 (3sf)
To find the least constant acceleration needed for takeoff from a 1.80 km runway, we can use the equation relating distance (s), initial velocity (u), final velocity (v), and acceleration (a):
v^2 = u^2 + 2as
In this case, the initial velocity is 0 km/h (since the jet is starting from rest) and the final velocity is 360 km/h. Converting these velocities to m/s:
u = 0 km/hr = 0 m/s
v = 360 km/hr = (360 * 1000) / 3600 = 100 m/s
The distance is 1.80 km = (1.80 * 1000) = 1800 m.
Substituting these values into the equation:
100^2 = 0^2 + 2a * 1800
Simplifying:
10000 = 3600a
Dividing both sides by 3600:
a = 10000 / 3600
a = 2.78 m/s^2
Therefore, the least constant acceleration needed for takeoff from a 1.80 km runway is approximately 2.78 m/s^2.
To find the least constant acceleration needed for takeoff, we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity (360 km/h)
u = initial velocity (0 km/h, as the jumbo jet starts from rest)
a = acceleration
s = distance (1.8 km)
First, we need to convert the velocities and distance to SI units (m/s and meters):
v_final = 360 km/h = (360 * 1000) / (60 * 60) m/s = 100 m/s
u = 0 km/h = 0 m/s
s = 1.8 km = 1800 m
Now we can rearrange the equation to solve for acceleration:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (100^2 - 0^2) / (2 * 1800)
a = 10000 / 3600
a ≈ 2.78 m/s^2
Therefore, the least constant acceleration needed for takeoff from a 1.80 km runway is approximately 2.78 m/s^2.