Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from rest to a take-off speed of 70 m/s over a distance of 940 m. Use of the kinematic equation (vx)2f=(vx)2i+2axΔx to find the acceleration of the plane, and then answer the following question. How long does it take the plane to reach its take-off speed?

Where [tex](vx)f^{2}[/tex] is the final velocity, [tex](vx)i^{2}[/tex] is the initial velocity, [tex]ax[/tex] is the acceleration and [tex]\Lambda x[/tex] is the distance traveled.

Equation (1) can be rewritten in terms of ax:

[tex](vx)f^{2} - (vx)i^{2} = 2ax \Lambda x[/tex]

[tex]2ax \Lambda x = (vx)f^{2} - (vx)i^{2}[/tex]

[tex]ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}[/tex] (2)

Since the plane starts from rest, its initial velocity will be zero ([tex](vx) = 0[/tex]):

Replacing the values given in equation 2, it is gotten:

[tex]ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}[/tex]

[tex]ax = \frac{4900m/s}{2(940m)}[/tex]

[tex]ax = \frac{4900m/s}{1880m}[/tex]

[tex]ax = 2.60m/s^{2}[/tex]

So, The acceleration of the plane is [tex]2.60m/s^{2}[/tex]

Now that the acceleration is known, the next equation can be used to find out the time:

[tex](vx)f = (vx)i + axt [/tex] (3)

Rewritten equation (3) in terms of t:

[tex]t = \frac{(vx)f - (vx)i}{ax}[/tex]

[tex]t = \frac{70m/s - 0m/s}{2.60m/s^{2}}[/tex]

[tex]t = 26.92s[/tex]

Hence, the plane takes 26.92 seconds to reach its take-off speed.