Help!! Question about average acceleration?

Question

An airport has runways only 190 m long. A small plane must reach a ground speed of 35 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 (s2 meaning s-squared).

Any help would be greatly appreciated!!!

Answer 1

You need two formula's:

V(t) = a t (1)
where V(t) = speed
a = acceleration
t = time

and

S(t) = 0.5 a t^2 (2)
where S(t) = distance

From (1) with V(t) = 35 m/s:
35 = a t
=> t = 35/a

From (2) with S(t) = 190 m:
190 = 0.5 a t^2
=> 190 = 0.5 a (35/a)^2 (substitute t)
=> 190 = 612.5/a
=> a = 612.5/190 = 3,22 m/s^2

Answer 2

You have a disagreement on final answer. I agree with those that said 3.22 m/s^2. Especially the formula used by DavidK93 and some others.

chaospersoni...'s mistake is at the step "to solve for t, t=d/v". The velocity is increasing as the plane goes down the runway. For v, the average velocity is needed, 35m/s is the final velocity. The average velocity would be 1/2 of the final. chaospersoni...'s result is therefore double.

Answer 3

The most useful formula for velocity in this context would be v^2 = v0^2 + 2ax. v is the final velocity, v0 is the initial velocity, a is acceleration (assumed to be constant), and x is the distance over which the acceleration takes place. The initial velocity here is 0, so the formula simplifies to v^2 = 2ax, which implies a = (v^2)/(2x). You know the required v (35 m/s) and you know the available x (190 m), so that's everything you need.

Answer 4

Its really easy Just use the equation
V2=U2+2*a*S

V=final Velocity
U=Starting Velocity
a=acceleration
s=displacement
So according to u r problem

U=0,V=35m/s,a=what u find,s=190m

so a=3.22m/s2

Answer 5

"Then I had to find the difference in position, divided by times squared, which I believe is also acceleration" The formula for distance traveled in time t under constant acceleration is d = (1/2)at^2. Rearranging that you get d/t^2 = (1/2)a.

Answer 6

Your question is very perceptive in that you refer to "average acceleration" while many think of acceleration as a constant like g for gravity (that in fact varies with distance from the center of the earth). A change in distance with time is velocity. A change in velocity with time is acceleration. But a change in acceleration with time is called "jerk" or "jolt." An elevator starting from the ground floor and going up many levels may need to use "jerk" to assure comfort and speed (smooth takeoff but increasingly faster and faster lift to save precious time). For the airplane experiencing uneven acceleration (because wind resistance and propeller effects are changing?) jerk may be present and it would be necessary to know its true nature. You may have to treat the "average acceleration" as a constant and use it to obtain a best estimate for a solution. Good luck.

Answer 7

as v=u+at with u=0 we have v=35 so at=35.

s=ut+0.5at2, so 190=0.5t35 so t=190/17.5.

now s=1/2at2 so 190=a0.5x(190/17.5)2
which gives a=3.22m/s2

Answer 8

a = (vf - vo)/t

where:

a - acceleration
vf - final velocity
vo - original velocity
t - time

from the problem we have the ff information:
d=190m
vf=35m/s
vo=0m/s
t=?

we know that distance = velocity *time or d=vt
to solve for t, t=d/v

t=(190m) / (35m/s)
t=5.43 s (rounded off)

now that we have the value fot t, we can use the formula above for acceleration

a = (vf - vo)/t
a= (35m/s-0m/s)/(5.43s)
a=6.45m/s^2

hope this help. 0_+

Answer 9

v2=u2+2as
u2=0.
so a=v2/2s=3.2237m/s2.

Answer 10

use v2=u2+2as