When a car driving up a hill with constant speed: I. its kinetic energy is decreasing. II. its potential energy is constant. III
1 answer:
A car driving up a hill at a constant speed experiences no change in its kinetic energy while it's potential energy increases with increasing height, thus none of the options are correct.
Understanding the concept
Consider a car moving up the hill at a constant speed as shown in the figure below. The following forces act on the car:
- N is the normal reaction force acting in an upward direction
- f_s is the static friction force exerted due to friction between the road and the tires of the car
- f_k is the rolling friction force in the direction opposing that of the tire
- mg is the force acting in a downward direction.
- θ is the angle of inclination.
Here as the car is moving up the hill at a constant speed, the net force exerted on the car is zero. Also, the kinetic energy of the car will not change as its velocity is constant and the potential energy will change with increasing height. Thus, none of the given options are correct.
Learn more about motion on an incline here:
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Answer:
a.
b.
c.
Explanation:
Modeling With Functions
Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by
For Carl Lewis's run at the 1987 World Championships, the values of a and b are
Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is
a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?
To compute the accelerations, we must find the function for a as the derivative of v
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For t=2
b. Find an expression for the distance traveled at time t.
The distance is the integral of the velocity, thus
To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates
Solving for C
Now we complete the equation for the distance
c. Find the time Lewis needed to sprint 100.0 m.
The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.
We are required to find the time at which the distance is 100 m, thus
Rearranging
We define an auxiliary function f(t) to help us find the value of t.
Let's try for t=9 sec
Now with t=9.9 sec
That was a real close guess. One more to be sure for t=10 sec
The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).