An airport has runways only 142 m long. Asmall plane must reach a ground speed of 45 m/s before it can become airborne. What average acceleration must the plane’s engines provide if it is to take off safely from its airport?Answer in units of m/s2.
i do not understand that formula bob
i got 0.317 what do you get cause mine is wrong
Thank you for the post
To find the average acceleration needed for the plane to take off safely, we can use the following formula:
Average acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
In this case, we need to find the change in velocity and we know the final velocity (v) is 45 m/s, but we don't know the initial velocity.
The initial velocity (u) can be found using the formula:
v = u + at
Rearranging the formula, we get:
u = v - at
Since the plane starts from rest (u = 0), we can rewrite the formula as:
0 = v - at
Solving for time (t), we get:
t = v / a
Now, we can substitute this value of time into the equation for the initial velocity:
u = 0 - at = -av
The plane covers a distance of 142 m, which can be written as:
Distance (s) = ut + (1/2)at^2
Since the plane starts from rest, we can simplify this equation to:
s = (1/2)at^2
Now, substituting the value of time (t) from above, we have:
s = (1/2)(a(v/a))^2
s = (1/2)v^2 / a
Rearranging this equation, we can solve for acceleration (a):
a = (1/2)v^2 / s
Plugging in the given values:
a = (1/2)(45 m/s)^2 / 142 m
Simplifying this expression:
a = 0.901 m/s^2
Therefore, the average acceleration the plane's engines need to provide for it to take off safely from its airport is 0.901 m/s^2.
Vf^2=Vi^2+2ad solve for a
ur stupid
Vf=45
Vi=0
d=142
45^2=2*a*142
solve for a.
Yep, yours is way wrong.