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)
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.
1 answer
1 answer