An object starts from rest with a constant acceleration of 8.0 m/s^2 along a straight line.
A)Find the distance traveled in the 5.0 s.
B)What is its speed just before it hits the ground?
A) Using the equation:
distance = initial velocity x time + (1/2) x acceleration x time^2
Since the object starts from rest, the initial velocity is 0. Therefore:
distance = (1/2) x acceleration x time^2
distance = (1/2) x 8.0 m/s^2 x (5.0 s)^2
distance = 100.0 m
Therefore, the object travels 100.0 meters in 5.0 seconds.
B) Using the equation:
final velocity^2 = initial velocity^2 + 2 x acceleration x distance
Again, the initial velocity is 0. Therefore:
final velocity^2 = 2 x acceleration x distance
final velocity^2 = 2 x 8.0 m/s^2 x 100.0 m
final velocity^2 = 1600.0 m^2/s^2
Taking the square root of both sides:
final velocity = 40.0 m/s
Therefore, the speed of the object just before it hits the ground is 40.0 m/s.
A plane is traveling eastward at an airspeed of 500&m/h. But a 90km/ wind is blowing southward. What is the direction of the plane relative to the ground?
To find the direction of the plane relative to the ground, we need to use vector addition. The airspeed of the plane is a velocity vector relative to the air, and the wind is another velocity vector. The direction of the plane relative to the ground is the resultant vector of these two velocity vectors.
Let's call the eastward direction the x-direction, and the southward direction the y-direction. Then the velocity vector of the plane with respect to the air is:
velocity_airplane = 500 km/h in the x-direction
And the velocity vector of the wind is:
velocity_wind = 90 km/h in the y-direction
To find the resultant velocity vector, we add the two vectors together using vector addition:
resultant_velocity = velocity_airplane + velocity_wind
resultant_velocity = 500 km/h in the x-direction + 90 km/h in the y-direction
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
|resultant_velocity| = sqrt((500 km/h)^2 + (90 km/h)^2)
|resultant_velocity| = 506.6 km/h
To find the direction of the resultant velocity, we can use trigonometry:
tan(theta) = (opposite/adjacent) = (90 km/h) / (500 km/h)
theta = arctan(0.18)
theta = 10.26 degrees
Therefore, the direction of the plane relative to the ground is 10.26 degrees east of due north.
To find the distance traveled in 5.0 s, we can use the formula for distance:
distance = (initial velocity × time) + ((1/2) × acceleration × time^2)
Given:
Initial velocity (u) = 0 m/s (as the object starts from rest)
Acceleration (a) = 8.0 m/s^2
Time (t) = 5.0 s
A) Calculating the distance traveled:
distance = (0 × 5.0) + ((1/2) × 8.0 × 5.0^2)
distance = 0 + ((1/2) × 8.0 × 25)
distance = 0 + (4.0 × 25)
distance = 0 + 100
distance = 100 m
Therefore, the distance traveled in 5.0 s is 100 m.
To find the speed just before it hits the ground, we can use the formula:
final velocity (v) = initial velocity (u) + (acceleration × time)
Given:
Initial velocity (u) = 0 m/s (as the object starts from rest)
Acceleration (a) = 8.0 m/s^2
Time (t) = ?
We need to find the time taken by the object to hit the ground.
The formula for the time of flight is:
t = (2 × initial velocity) / acceleration
Given:
Initial velocity (u) = 0 m/s (as the object starts from rest)
Acceleration (a) = 8.0 m/s^2
B) Calculating the time of flight:
t = (2 × 0) / 8.0
t = 0 / 8.0
t = 0 s
The time of flight is 0 s, which means the object hits the ground instantaneously.
To find the final velocity:
final velocity (v) = initial velocity (u) + (acceleration × time)
v = 0 + (8.0 × 0)
v = 0 m/s
Therefore, the speed just before it hits the ground is 0 m/s.