1. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration -5.20 m/s2 for 4.45 s, making straight skid marks 62.5 m long ending at the tree. With what speed does the car then strike the tree?
2. A jet plane lands with a speed of 80.0 m/s and can accelerate with a maximum magnitude of 7.00 m/s2 as it comes to rest.
(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest?
(b) Can this plane land on a small tropical island airport where the runway is 0.800 km long?
yes or no
(c) What is the minimum distance the plane requires to land? (in km)
1. You know the distance. Find the terminal velocity.
Vf^2=Vo^2-2ad
2. Nuts to the problem statement. Planes have to deaccelerate to stop.
Vf=Vo-a*t solve for time t.
distance=Vo*tmin -1/2 a tmin^2
1. Well, it seems like the driver took "tree-blocking-the-road" quite literally and decided to give it a tight hug. Ouch! To answer your question, let's calculate the speed of the car when it embraces the tree of love. We know that the initial velocity of the car is 0 m/s because it started from rest, and the acceleration is -5.20 m/s². So, we can use the formula:
Vf = Vi + at
Plugging in the values, we have:
Vf = 0 + (-5.20 m/s²) × 4.45 s
Vf = -23.14 m/s
Negative speed, huh? That's as if the car is moonwalking into the tree. Let's take the absolute value to make it positive because we can't have negative speeds.
Therefore, the speed at which the car lovingly embraces the tree is approximately 23.14 m/s.
2. Ah, airplanes, those majestic birds that just happen to be powered by jet engines instead of feathers. Okay, let's tackle these questions one by one.
(a) The plane, with a landing speed of 80.0 m/s and a maximum acceleration of 7.00 m/s², wants to come to a graceful rest. To determine the minimum time interval required for this feat, let's use a simple equation:
Vf = Vi + at
Since the final velocity is 0 m/s (rest), the initial velocity is 80.0 m/s, and the acceleration is -7.00 m/s² (negative because it's deceleration), we can rejiggle the equation:
0 = 80.0 m/s + (-7.00 m/s²) × t
Now solving for time:
t = (80.0 m/s) / (7.00 m/s²)
t ≈ 11.43 s
So, the airplane needs approximately 11.43 seconds to gracefully come to rest after touchdown.
(b) Ah, the small tropical island airport. Sun, sand, and... a short runway. The runway length is 0.800 km, or 800 meters. To determine if our jet plane can safely land on this cozy strip, we'll need to calculate the distance it requires to come to rest. We can use another equation for this:
Vf² = Vi² + 2ad
Since the final velocity is 0 m/s, the initial velocity is 80.0 m/s, and the acceleration is -7.00 m/s², we can plug them in:
0 = (80.0 m/s)² + 2(-7.00 m/s²) × d
Now it's time to solve for distance:
d = [(80.0 m/s)²] / [2(-7.00 m/s²)]
d ≈ 3200 m
Uh-oh! The plane needs a minimum distance of approximately 3200 meters to land safely. The runway falls a little short, so our plane will have to find another beachy destination to touchdown on.
(c) The minimum distance the plane requires to land is approximately 3200 meters or 3.2 kilometers. No palm tree obstacles, please!
1. To find the speed at which the car strikes the tree, we can use the equation of motion:
v = u + at
Where:
- v is the final velocity (unknown)
- u is the initial velocity (0 m/s)
- a is the acceleration (-5.20 m/s²)
- t is the time (4.45 s)
Using the given values, we can calculate the final velocity:
v = 0 + (-5.20 × 4.45) m/s
v = -23.14 m/s
The negative sign indicates that the car is moving in the opposite direction, towards the tree. Therefore, the magnitude of the velocity is:
|v| = |-23.14| m/s
|v| = 23.14 m/s
Therefore, the car strikes the tree with a speed of 23.14 m/s.
2.
(a) To find the minimum time interval needed before the plane can come to rest, we can use the equation of motion:
v = u + at
Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (80.0 m/s)
- a is the acceleration (-7.00 m/s², as it slows down)
- t is the time (unknown)
Using the given values, we can calculate the time interval:
0 = 80.0 + (-7.00 × t) m/s
-80.0 = -7.00t
t = (-80.0) / (-7.00) s
t ≈ 11.43 s
Therefore, the minimum time interval needed for the plane to come to rest is approximately 11.43 seconds.
(b) To determine if the plane can land on the small tropical island airport runway, we need to calculate the distance it would travel before coming to rest. Since it starts with an initial velocity of 80.0 m/s and comes to rest at an acceleration of -7.00 m/s², we can use the equation:
v² = u² + 2as
Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (80.0 m/s)
- a is the acceleration (-7.00 m/s²)
- s is the distance (unknown)
Rearranging the equation, we have:
s = (v² - u²) / (2a)
Substituting the given values:
s = (0 - (80.0)²) / (2 × (-7.00)) m
s = (-6400.0) / (-14.00) m
s ≈ 457.14 m
Therefore, the plane would require a distance of approximately 457.14 meters to come to rest. Since the runway length is 0.800 km (800 meters), the plane can land on the small tropical island airport runway.
(c) The minimum distance the plane requires to land is approximately 457.14 meters, which is equivalent to 0.45714 km.
To solve these problems, we can use the equations of motion.
For the first question:
1. We are given the acceleration of the car as -5.20 m/s^2 and the time it takes to come to a stop as 4.45 s. We need to find the initial speed of the car before braking.
2. The equation we will use is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
3. Rearranging the equation, we get u = v - at.
4. Since the car comes to a stop, the final velocity v is 0.
5. Plugging in the values, we have u = 0 - (-5.20 m/s^2 * 4.45 s).
6. Calculating that, we get u = 23.14 m/s.
7. So, the car strikes the tree with a speed of 23.14 m/s.
For the second question:
(a) To find the minimum time interval needed for the plane to come to rest, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
1. We are given the final velocity v as 0 m/s and the maximum acceleration as 7.00 m/s^2.
2. Rearranging the equation, we get t = (v - u) / a.
3. Plugging in the values, we have t = (0 - 80.0 m/s) / (-7.00 m/s^2).
4. Calculating that, we get t = 11.43 s.
5. So, the minimum time interval needed before the plane can come to rest is 11.43 seconds.
(b) To determine if the plane can land on the small tropical island airport, we need to compare the length of the runway to the minimum distance the plane requires to land.
1. We are given the length of the runway as 0.800 km.
2. The minimum distance required to land is equal to the distance covered when the plane comes to rest.
3. Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange to solve for the distance.
4. Rearranging the equation, we get s = (v^2 - u^2) / (2a).
5. Plugging in the values, we have s = (0^2 - 80.0^2 m/s) / (2 * -7.00 m/s^2).
6. Calculating that, we get s = 163.265 m.
7. Converting 163.265 m to km, we get 0.163 km.
8. The length of the runway is 0.800 km, which is greater than 0.163 km.
9. Therefore, yes, the plane can land on the small tropical island airport.
(c) The minimum distance required to land is determined by the calculation in step 6 above: 0.163 km.