There's 2.
1) A 1200-kg airplane starts from rest and, with a constant average forward acceleration of magnitude 5.0m/s^2, reaches its take off speed in 9.00 s. The runway is 300 m long. What is the planes takeoff speed?
2) A boy is attempting to swim directly across a river. He is able to swim at a speed of .500 m/s relative to the water. the river is 25.0m wide and the boy end up at 50.0m downstream from his starting point. How fast is the current of the river?
Please post one question at a time and show your work.
1) The acceleration rate times the time spend taking off is the takeoff speed. Two of the numbers you were given are not needed to solve the problem. Takeoff does not require the full runway length. It would be dangerous if it did.
Oh I see. So 45m/s
To solve these problems, we will use the principles of kinematics and vector addition.
1) To find the plane's takeoff speed, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (takeoff speed)
u = initial velocity (0 m/s as the plane starts from rest)
a = acceleration (5.0 m/s^2)
s = distance (300 m)
Substituting the given values into the equation, we get:
v^2 = 0^2 + 2(5.0 m/s^2)(300 m)
v^2 = 2(5.0 m/s^2)(300 m)
v^2 = 3000 m^2/s^2
v = sqrt(3000 m^2/s^2)
v ≈ 54.77 m/s
Therefore, the plane's takeoff speed is approximately 54.77 m/s.
2) To find the speed of the river current, we can use the concept of vector addition. The boy's actual velocity across the river will be the combination of his swimming speed and the river's current speed.
Let's assume the current speed is denoted by v_c.
The boy's velocity across the river can be represented by the equation:
v_b = sqrt(v_w^2 + v_c^2)
Where:
v_b = velocity of the boy across the river (unknown)
v_w = velocity of the boy relative to the water (0.500 m/s, as given in the problem)
v_c = velocity of the river current (unknown)
We also know that the boy ends up 50.0 m downstream from his starting point. Since the boy's velocity across the river is perpendicular to the river's current, we can use basic trigonometry to solve for v_c.
Using the formula v_c = d / t, where d = 50.0 m and t = 25.0 s (as the boy ends up 50.0 m downstream from his starting point in 25.0 s), we can find:
v_c = 50.0 m / 25.0 s
v_c = 2.0 m/s
Therefore, the speed of the river current is 2.0 m/s.