Starting from a location with position vector r1,x =−17.5 m and r1,y=23.1 m , a rabbit hops around for 10.7 seconds with average
1 answer:
The kinematics of the uniform motion allows us to find the final position vector
r = (-41.575 i + 42.253 j) m
Given parameters
- the starting position x = -17.5 m y = 23.1 m
- jump time t = 10.7 s
- The average velocities vₓ = -2.25 m / s and v_y = 1.79 m / s
to find
- the final position
The uniform motion occurs when the velocity of the bodies is constant, in this case the relationship can be used for each axis
v =
x = x₀ + v t
Where vₓ it is the velocity, x the displacement, x₀ the initial position and t the time
Let's set a reference system with the horizontal x-axis. Regarding which we carry out the measurements
X axis
we look for the final position
x = x₀ + vₓ t
x = -17.5 -2.25 10.7
x = -41.575 m
Y Axis
we look for the final position
y = y₀ + v_y t
y = 23.1 + 1.79 10.7
y = 42.253 m
In conclusion, using the kinematics of uniform motion, find the final position vector
r = (-41.575 i + 42.253 j) m
learn more about uniform motion here:
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31) The head of each arrow represents the direction of each force
32) The length of each arrow represents the magnitude of each force
33) The net force on the 1-kg object is to the left
34) To balance the forces on the object, we must add a force acting to the right
35) The acceleration of the 1-kg object is twice the acceleration of the 2-kg object
36) The magnitude of the acceleration increases, and the direction changes
Explanation:
31-32)
A free-body diagram is a diagram representing the forces acting on a body.
In a free-body diagram, each force is represented with an arrow, where:
- The length of the arrow is proportional to the magnitude of the force
- The direction of the arrow represents the direction of the force
Therefore:
- The head of each arrow represents the direction of each force
- The length of each arrow represents the magnitude of each force
33)
In the free-body diagram of the 1-kg object, we observe that the arrow on the right is longer than the arrow on the left. This means that the force acting from the right (and therefore, acting TO the left) is stronger than the force acting from the left (which is therefore acting TO the right).
Therefore, the net force acting on the 1-kg object is to the left, since the force acting to the left is stronger.
34)
In order to balance the forces acting on the 1-kg, we should apply an additional force such that the net force acting on the object is zero.
In part 33), we said that the net force acting on the 1-kg object is to the left. Therefore, in order to produce a net force of zero, we must add another force acting to the right, in order to balance the current net force pointing to the left.
35)
First of all, by looking at the two free-body diagrams, we notice that the magnitudes of the forces acting on the 1-kg and 2-kg objects are the same. This means that the net force acting on the two objects is the same.
Now we can apply Newton's second law of motion to the two cases; the law states that the net force acting on an object is equal to the product between its mass and its acceleration:
where is the net force, m is the mass, a is the acceleration
For the object of 1-kg, we have:
where
For the object of 2-kg, we have:
where
We said that the net force on the two objects is the same, so we can combine the two equations:
Therefore, the acceleration of the 1-kg object is twice the acceleration of the 2-kg object.
36)
If an upward force is applied to the 1-kg object, the acceleration will increase in magnitude and it will change in direction.
In fact, let's call the total net force acting in the horizontal direction on the 1-kg object. Let's call
the upward additional force applied. This means that now the object will have two components of the acceleration:
, horizontally (to the left)
, vertically (upward)
So, the magnitude of the net acceleration will be
which is larger than the original acceleration (). Also, since there is a component of the acceleration in the vertical direction, the final direction of the acceleration
will be between left and upward.
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