A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50x^5 kg. Its speed is 27.0 m/s, and the net braking force is 4.30x10^5 N. (a)what is its speed 7.50 s later? (b)how far has it traveled in this time?
Use the braking force and F = ma to get the deceleration rate, a.
(a) After t= 7.5 seconds, the velocity will be
V(7.5) = Vo - a t
= 27.0 m/s - a*7.5
(b) Distance travelled at t = 7.5s
= (Average velocity)* time
= (1/2)*(Vo + V(7.5)) * 7.5s
The answer is found with your teacher. Go ask your teacher
To answer these questions, we need to use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
(a) To find the speed of the jetliner 7.50 seconds later, we can use the equation:
final velocity = initial velocity + (acceleration × time)
First, let's find the acceleration using Newton's second law:
Net force = mass × acceleration
Rearranging the equation to solve for acceleration:
acceleration = Net force / mass
Substituting the given values:
acceleration = 4.30x10^5 N / 3.50x10^5 kg
Now, we can determine the final velocity using the equation:
final velocity = 27.0 m/s + (acceleration × 7.50 s)
(b) To find the distance traveled by the jetliner in 7.50 seconds, we can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
Substituting the given values:
distance = 27.0 m/s × 7.50 s + 0.5 × acceleration × (7.50 s)^2
Let's calculate both the final velocity and the distance traveled using the given values.