A jet plane lands with a speed of $$ 150 \mathrm{~m} / \mathrm{s} $$ and can accelerate at a maximum rate of $$ 7.00 \mathrm{~m} / \mathrm{s}^{\wedge} 2 $$ as it comes to rest. a) From the instant the plane touches the runway, what is the minimum time it needs before it can come to rest? b) Can this plane land at a small tropical island airport where the runway is 0.900 kn long?

Question

UpStudy AI · Accepted Answer

**(a)**  
Given the initial speed $$ v_0 = 150 \, \mathrm{m/s} $$ and the maximum deceleration $$ a = 7.00 \, \mathrm{m/s^2} $$, we can use the kinematic equation for velocity under constant acceleration:
$$
v = v_0 - at
$$
To come to rest, the final speed $$ v = 0 $$. Setting the equation to zero gives:
$$
0 = 150 - 7t
$$
Solving for $$ t $$:
$$
t = \frac{150}{7} \approx 21.4 \, \mathrm{s}
$$

**(b)**  
First, convert the runway length from kilometers to meters:
$$
0.900 \, \mathrm{km} = 900 \, \mathrm{m}
$$
We now calculate the stopping distance using the kinematic formula:
$$
v^2 = v_0^2 - 2ad
$$
Again with $$ v = 0 $$ (when the plane stops), rearrange to solve for $$ d $$:
$$
0 = 150^2 - 2(7)d \quad \Longrightarrow \quad d = \frac{150^2}{2 \times 7}
$$
Calculating:
$$
d = \frac{22500}{14} \approx 1607 \, \mathrm{m}
$$

Since $$ 1607 \, \mathrm{m} $$ (the minimum stopping distance) is greater than the available runway length $$ 900 \, \mathrm{m} $$, the plane cannot land safely on that runway.

**Answer:**  
(a) The minimum time is approximately $$ 21.4 \, \mathrm{s} $$.  
(b) No, the plane would require a runway of approximately $$ 1607 \, \mathrm{m} $$, which is longer than the $$ 900 \, \mathrm{m} $$ runway available.
**(a)** The plane needs about 21.4 seconds to come to rest. **(b)** No, the plane cannot land safely on the 900-meter runway because it requires approximately 1607 meters to stop.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Physics Questions