A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.5 x 10 exponent 5 kg, its speed is 27.0 m/s and the net braking force is 4.30 x 10 exponent 5 N, (A) what id its speed 7.50 s later? (b) how far has it traveled in this time?
First calculate the deceleration rate:
a = F/m = (4.30*10^5 N)/(3.5*10^5 kg)
= 1.229 m/s^2
(a) At that deceleration rate, the speed 7.50 s later is reduced by 7.5*1.229 = 9.2 m/s, making the speed 17.8 m/s at that time.
(b) For the distance traveled, multiply the AVERAGE speed by 7.50 s.
22.4 m/s * 7.5 s = ___ m
A) Well, let's see what happens. The net force acting on the 747 is the braking force, which is constant. We can use Newton's second law (Force = mass x acceleration) to find the acceleration.
Force = mass x acceleration
Rearranging the equation, we get:
Acceleration = Force / mass
Acceleration = (4.30 x 10^5 N) / (3.5 x 10^5 kg)
Calculating this, we find the acceleration is approximately 1.23 m/s^2.
Now, since the force is acting in the opposite direction to the motion of the jetliner, it will decelerate. Using the equation of motion (v = u + at), we can find the velocity (v) of the jetliner 7.50 seconds later.
Given:
Initial velocity, u = 27.0 m/s
Acceleration, a = -1.23 m/s^2 (negative because it's deceleration)
Time, t = 7.50 s
v = u + at
v = 27.0 m/s + (-1.23 m/s^2)(7.50 s)
Calculating this, we find the final velocity (v) of the jetliner after 7.50 seconds is approximately 17.8 m/s.
Therefore, the speed of the jetliner 7.50 seconds later is approximately 17.8 m/s.
B) To find the distance traveled (s) in this time, we can use another equation of motion:
s = ut + (1/2)at^2
Given:
Initial velocity, u = 27.0 m/s
Acceleration, a = -1.23 m/s^2 (negative because it's deceleration)
Time, t = 7.50 s
s = (27.0 m/s)(7.50 s) + (1/2)(-1.23 m/s^2)(7.50 s)^2
Calculating this, we find the distance traveled (s) by the jetliner in 7.50 seconds is approximately 177.2 meters.
Therefore, the jetliner has traveled approximately 177.2 meters in this time.
Hope that adds a bit of amusement to your physics problem!
To find the speed of the 747 jetliner 7.50 seconds later, we need to use the equation of motion that relates acceleration, initial velocity, time, and final velocity:
v = u + at
where:
v = final velocity (unknown)
u = initial velocity (given as 27.0 m/s)
a = acceleration (unknown)
t = time (given as 7.50 s)
To find the acceleration, we can use Newton's second law of motion:
F = ma
where:
F = net braking force (given as 4.30 x 10^5 N)
m = mass of the jetliner (given as 3.5 x 10^5 kg)
a = acceleration (unknown)
Rearranging the equation to solve for acceleration:
a = F / m
Plugging in the values:
a = (4.30 x 10^5 N) / (3.5 x 10^5 kg)
Calculating the acceleration:
a = 1.23 m/s^2
Now, we can use this acceleration value in the first equation to find the final velocity:
v = u + at
v = 27.0 m/s + (1.23 m/s^2) * 7.50 s
Calculating the final velocity:
v = 27.0 m/s + 9.23 m/s
v ≈ 36.23 m/s
Therefore, the speed of the 747 jetliner 7.50 seconds later is approximately 36.23 m/s.
To find the distance traveled, we can use another equation of motion:
s = ut + 0.5at^2
where:
s = distance (unknown)
u = initial velocity (given as 27.0 m/s)
t = time (given as 7.50 s)
a = acceleration (1.23 m/s^2)
Plugging in the values:
s = (27.0 m/s) * (7.50 s) + 0.5 * (1.23 m/s^2) * (7.50 s)^2
Calculating the distance traveled:
s = 202.5 m + 40.96 m
s = 243.46 m
Therefore, the 747 jetliner has traveled approximately 243.46 meters in 7.50 seconds.
To answer these questions, we need to use the concepts of Newton's second law of motion and the equations of motion.
(A) To find the speed of the jetliner 7.50 seconds later, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity is 27.0 m/s, the acceleration can be calculated using Newton's second law:
F = ma
Where:
F = net force
m = mass
Rearranging the equation, we get:
a = F/m
Now we can substitute the values:
a = (4.30 × 10^5 N) / (3.5 × 10^5 kg)
a ≈ 1.23 m/s^2
Now we can use the equation of motion to find the final velocity:
v = u + at
v = 27.0 m/s + (1.23 m/s^2) * (7.50 s)
v ≈ 36.98 m/s
Therefore, the speed of the jetliner 7.50 seconds later is approximately 36.98 m/s.
(B) To find the distance traveled by the jetliner in this time, we can use a different equation of motion:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity
a = acceleration
t = time
Using the previously calculated acceleration (a ≈ 1.23 m/s^2) and the given time (t = 7.50 s), we have:
s = (27.0 m/s) * (7.50 s) + (1/2) * (1.23 m/s^2) * (7.50 s)^2
s ≈ 202.31 m
Therefore, the jetliner has traveled approximately 202.31 meters in this time.