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Montano1993 [528]
3 years ago
8

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?
Physics
1 answer:
kodGreya [7K]3 years ago
5 0

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

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

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