The average velocity for a trip has a positive value. Is it possible for the instantaneous velocity at any point during the trip to have a negative value? (I'm thinking yes because derivatives can be negative but I'm not sure).
also
A ball is dropped from rest from the top of a building and strikes the ground with a speed Vf. From ground level, a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is Vo=Vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building, above the halfway point, or below the halfway point. Give your reasoning. (I'm thinking the second ball must be traveling faster if it starts at the fastest speed of the first ball, so they'd cross above the halfway point. Is that right?)
Yes to the first question. They must be talking about motion along one axis. You can go backwards for a while (negative velocity) and still get you your destination.
As for the second question
Y1 = H - (g/2)t^2
Vf = sqrt (2gH)
Y2 = Vf*t - (g/2)t^2
When Y1 = Y2,
Vf*t = H
Vf^2 t^2 = H^2
2 g H t^2 = H^2
gt^2 = H/2
Y1 = Y2 = 3H/4
Yes, you are right, and your reasoning led you to the correct answer right away
Yes, it is possible for the instantaneous velocity at any point during a trip to have a negative value. The average velocity is calculated by dividing the total displacement by the total time taken. It represents the overall velocity of the trip. However, the instantaneous velocity represents the velocity at a particular instant in time and can have a different value. The instantaneous velocity is the limiting value of the average velocity as the time interval approaches zero. Therefore, at any given moment, the instantaneous velocity can have different signs (positive or negative) depending on the direction of motion.
Regarding the second question, when the first ball is dropped from rest, it will accelerate downward due to gravity. The second ball, thrown upward at the same instant, will initially have the same speed as the first ball when it eventually strikes the ground (Vf). However, while the first ball has only been accelerating downward, the second ball experiences both upward acceleration from the throw and later downward acceleration due to gravity.
Since the second ball starts with an initial velocity equal to the speed with which the first ball will strike the ground, it will reach the same maximum height as the first ball but take longer to come back down. Remember that the time taken for the ball to reach its maximum height is the same as the time taken to fall from the maximum height to the ground (assuming no air resistance).
Therefore, the second ball will cross paths with the first ball above the halfway point of the building. This is because the second ball needs more time to reach its maximum height and start descending, while the first ball is already falling down. So, the two balls will cross paths at a height above the halfway point.
For the first question, yes, it is possible for the instantaneous velocity at any point during the trip to have a negative value. In order to understand this, let's first clarify the difference between average velocity and instantaneous velocity.
Average velocity is defined as the total displacement divided by the total time taken. It tells us the overall direction and magnitude of motion over a given trip or interval. It can have a positive value, indicating overall motion in a particular direction.
On the other hand, instantaneous velocity refers to the velocity of an object at a specific instant in time. It is determined by calculating the derivative of the displacement with respect to time. The derivative represents the rate of change of displacement with respect to time, or in other words, the slope of the displacement-time graph at a particular point.
Since the derivative can be positive or negative, the instantaneous velocity can have a negative value even if the average velocity is positive. This occurs when the displacement is negative over a very short interval or when the object changes its direction momentarily.
Moving on to the second question, we want to determine whether the balls will cross paths at half the height of the building, above the halfway point, or below the halfway point when the second ball is thrown straight upward at the same instant that the first ball is dropped.
Assuming no air resistance, both balls will experience the same acceleration due to gravity, which is constant near the Earth's surface. When the first ball is dropped, it accelerates downward due to gravity. At the moment the second ball is thrown upward, it will have an initial velocity equal to the final velocity of the first ball when it hits the ground.
To determine the trajectory of the second ball, we need to consider the motion of both balls. The first ball is accelerating downward, while the second ball is initially moving upward. The speed of the second ball is the same as the speed with which the first ball will eventually hit the ground.
Since the second ball is starting with an initial velocity equal to the final velocity of the first ball, it would take the same amount of time for the second ball to reach its peak height as it would for the first ball to hit the ground.
Therefore, the second ball will reach its peak height at half the time it takes for the first ball to hit the ground. This means that the second ball will cross paths with the first ball exactly at half the height of the building.