An airplane accelerates down a runway at 3.20 m/s_2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff?

2 Answers
A08
Feb 8, 2016

Distance traveled #s=1721.344m#
or #approx 1.721 km#

Explanation:

Aide memoires

There are three equations for 1D motion involving constant acceleration.

  1. #v=u+a t#
  2. #s=ut+1/2at^2#
  3. #v^2-u^2=2as#
    where symbols have meanings ascribed to each.

In the problem given quantities are acceleration #a#, time #t#. And we need to find distance #s# traveled.

Inspection reveals that equation 2 is the most suitable as it has all three quantities of interest in it except initial velocity #u#.

Answer starts from here #->#
We know that

#s=ut+1/2at^2#
where #s# is the distance traveled, #a# is constant acceleration experienced by an object, #u# is its initial velocity and #t# time of travel under this acceleration.

Now we assume that the airplane starts from rest on the runway. Therefore, first term on the RHS is zero. Inserting given values in the equation
#s=1/2xx 3.20xx32.8^2#

#s=1721.344m#

A08
Feb 8, 2016

Supplementary answer below.

Explanation:

(i) Let #u# be the initial velocity of any object which is subjected to constant acceleration #a# for time #t#,

Acceleration is rate of change of velocity,

#:.# Final velocity #v# after time #t# is given as
#=# Initial velocity #u# + acceleration #axx# time #t# for which acceleration acted on the object.

You obtain first equation.
#color(blue)(v=u+at)#

(ii) Once we know the initial velocity and final velocity
Average velocity #color(green)(v_(avg)=(u+v)/2)#
Distance #s# traveled in time #t= #Average velocity# xx #time,# s=(u+v)/2 xx t# ....... (A)
#=>(ut)/2+(vt)/2 #
Insert the value of #v# from first equation
#=>(ut)/2+((u+at)t)/2 #
#=>(ut)/2+(ut)/2+(atxxt)/2 #

You obtain second equation.

#color(blue)(s=ut+1/2at^2) #

(iii) Eliminate time #t# from the first and equation (A), to obtain third equation.

#v=u+at#,

taking #u# to the other side of equation to find the value of #t#

#t=(v-u)/a#
Insert #t# in (A) above

#s=(u+v)/2 xx (v-u)/a#

You get the third one as #(x+y)(x-y)=x^2-y^2#
#color(blue)(v^2-u^2=2as)#

Above equation is direct outcome of the Law of Conservation of Energy .

Change in kinetic energy #="Final KE"-"Initial KE"#
#=>#Change in kinetic energy #=1/2mv^2-1/2m u^2# ..........(B)

Work done by a constant force #vecF# to move an object of mass #m# through a distance #vecs# is given by

#W=vecFcdotvecs#

Rewriting it in terms of acceleration we get

#W=(mveca)cdot vecs#

As direction of force/acceleration and motion is same, we get

#W=mas# ...........(C)

Using Law of Conservation of energy

Change in kinetic energy #=W#
#:.1/2mv^2-1/2m u^2=mas#
#=>color(blue)(v^2-u^2=2as)#

Another thing one must remember that even though it is not written explicitly, in all above equations, symbols #a,u,v and s# are vectors.

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