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

To find the acceleration of the plane, we can use the given kinematic equation (vx)2f=(vx)2i+2axΔx. The acceleration of the plane is approximately 2.61 m/s². The time it takes the plane to reach its take-off speed is approximately 26.82 seconds.

Step-by-step explanation:

To find the acceleration of the plane, we can use the given kinematic equation (vx)2f=(vx)2i+2axΔx. The initial velocity(vx)i is 0 since the plane starts from rest, and the final velocity (vx)f is 70 m/s. The distance Δx is 940 m. Plugging in these values, the equation becomes 70^2=0+2a(940), which simplifies to 4900=1880a.

To find the acceleration, solve for a by dividing both sides of the equation by 1880. The acceleration is a=4900/1880=2.61 m/s².

Next, to find the time it takes the plane to reach its take-off speed, we can use the equation vf = vi + at, where vf is the final velocity (70 m/s), vi is the initial velocity (0 m/s), a is the acceleration (2.61 m/s²), and t is the time. Plugging in these values, the equation becomes 70 = 0 + 2.61t.

Solve for t by dividing both sides of the equation by 2.61. The time is t = 70 / 2.61 ≈ 26.82 seconds.

Answer 2

`ax = 2.60m/s^(2)`,
`t = 26.92s`

Step-by-step 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 = ((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 = ((70m/s)^(2) - (0m/s)^(2))/(2(940m))`

`ax = (4900m/s)/(2(940m))`

`ax = (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 = ((vx)f - (vx)i)/(ax)`

`t = (70m/s - 0m/s)/(2.60m/s^(2))`

`t = 26.92s`

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