Answer:
The work done by Joel is greater than the work done by Jerry.
Explanation:
Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

Where:
- Work, measured in joules.
- Force exerted on the object, measured in newtons.
- Travelled distance by the object, measured in meters.
During the first 10 minutes, the net work exerted on the object is zero. That is:





In exchange, the net work in the next 5 minutes is the work done by Joel on the object:


Hence, the work done by Joel is greater than the work done by Jerry.
<h2>
Person must have 8.18 m/s to catch the ball</h2>
Explanation:
Consider the vertical motion of ball
We have equation of motion s = ut + 0.5at²
Initial velocity, u = 12 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = -25 m
Substituting
-25 = 12 x t + 0.5 x -9.81 x t²
4.905 t² -12t - 25 = 0
t = 3.79 sec
Ball hits ground after 3.79 seconds.
So person need to cover 31 m in 3.79 seconds
Consider the horizontal motion of person
We have equation of motion s = ut + 0.5at²
Initial velocity, u = ?
Acceleration, a = 0 m/s²
Displacement, s = 31 m
Time, t = 3.79 seconds
Substituting
31 = u x 3.79 + 0.5 x 0 x 3.71²
u = 8.18 m/s
Person must have 8.18 m/s to catch the ball
Answer:
The answer is below
Explanation:
We are to check if the statement is true of false. If it is false, we correct the statement.
Solution:
Acceleration is the time rate of change of velocity. It is the ratio of the change in velocity to the change in time. The acceleration can be gotten from a velocity time graph by finding the slope of the graph.
The x coordinate represent the time and the y coordinate velocity.
5) Graph A passes through the point (0, 0) and (4, 24). Therefore the acceleration (slope) is:
Acceleration = 
This is correct.
6) Graph B is a straight line of 12 m/s. It passes through (0, 12) and (4, 12). Hence:
Acceleration = 
This is false.
Therefore the acceleration of graph B is 0 m/s².