Complete Question
A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?
Answer:
The values is
Explanation:
From the question we are told that
The distance it travels eastward is
The distance it travel towards the southeast is
The distance it travel towards the south is
Let x-axis be east
y-axis south
z-axis into the ground
The angle made between east and south is
The displacement toward x-axis is
The displacement toward the y-axis is
Now the overall displacement of the rat is mathematically evaluated as
B
Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is
where a = 1.2 km north
and b = 1.6 km east
magnitude = 2 km
Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.
Answer:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.
As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]