Question

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?

Answer 1

Answer:

$ax = 2.60m/s^{2}$, $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

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

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

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$  (2)

Since the plane starts from rest, its initial velocity will be zero ($(vx) = 0$):

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

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

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

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

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

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

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