Kinematics Flashcards
If an object falling at a speed of 50 m/s experiences an air resistance of 40 N, what will be the air resistance if the same object is falling at 150 m/s?
Air resistance is proportional to the square of velocity. That means that if the velocity triples, the air resistance will be increased 9-fold. 9 x 40 N results in a new air resistance of 360 N.
True or False: If a basketball is thrown up into the air, it will hit the ground at the same time that it would if it was simply dropped.
False
If a cannonball is fired at 100 m/s at a 30 degree angle above the horizontal, what is the initial vertical velocity? (Note that cos 30 = 0.87 and sin 30 = 0.5.)
Because velocity in the y-direction is equal to sinθ times the overall force, we can do the following calculations: vy = vi sinθ vy = (100 m/s) sin(30) vy = 50 m/s
How long will it take for a ball to hit the ground if it is thrown vertically up in the air at a velocity of 50 m/s?
10 seconds: For the first half of the ball’s flight, the initial velocity will be 50 m/s and the final velocity will be equal to 0 m/s (as it reaches the peak of its flight). Acceleration will be g, and will be causing the ball to slow down, so we can use the value of -10 m/s2. vf = vi + at (0)= (50) + (-10) t -50 = -10t t = 5 seconds. However, that is only the first half of the flight! The ball must come back down, which will take the same amount of time as it took to go up. Thus, the total flight time will be 10 seconds.
A ball is thrown so that it has an initial horizontal velocity of 30 m/s and an initial vertical velocity of 6 m/s. If it is in the air for 6 seconds, how far will it have be from its point of origin?
This question is much simpler than it looks. The velocity in the x-direction will never change. Since we know that the ball is in the air for 6 seconds, we can calculate the distance traveled by multiplying the horizontal velocity by time: d = vt = (30 m/s)(6 s) = 180 meters
A car starts from rest and accelerates at 10 m/s2 for 4 seconds. How far does it travel?
80 m To solve this problem, we must use the formula d = vi(t) + (1/2) at2 . Plugging in the values given yields: d = vi(t) + (1/2) at2 d = (0 m/s)(4 s) + (1/2)(10 m/s2) (4 s)2 d = 0 + (1/2)(10)(16) d = 80 meters
A motorcyclist initially traveling at 70 m/s accelerates at a constant acceleration to a speed of 130 m/s in 10 seconds. How far did he travel?
1000 m To solve this problem, we can use the formula d = (½)(vi + vf)(t). Plugging in the values given yields: d = (½) (70 m/s + 130 m/s)(10 s) d = (½)(200 m/s)(10 s) d = 1000 m
If a car accelerates from rest to a velocity of 30 m/s over 50 meters, what is the acceleration of the car, assuming it remains constant?
To solve this problem, we can use the formula vf2 = vi2 + 2ad. Plugging in the given values yields: vf2 = vi2 + 2ad (30 m/s)2 = (0 m/s)2 + 2a(50 m) 900 = 100a 9 m/s2 = a
How fast would a ball be falling if it is dropped and allowed to accelerate for 4 seconds? Assume no air resistance.
We can use the formula vf = vi + at to solve this problem. Plugging in the given values yields: vf = vi + at vf = 0 m/s + (10 m/s2)(4 s) vf = 40 m/s
An object is falling at its terminal velocity. If we increase the object’s mass by a factor of nine without affecting its aerodynamics, what will happen to that terminal velocity?
It will triple. At terminal velocity, the force of gravity must be equal to the force of air resistance. If the mass is increased by a factor of nine, then the gravitational force would increase by a factor of nine, since the force of gravity is equal to mg. In order for the force of air resistance to also increase by a factor of nine, the velocity would have to be tripled. This is because the force of air resistance is proportional to the square of velocity.
If a juggler throws a ball straight up at a speed of 25 m/s, how long will it take to come back down?
5 seconds. If we look at the first half of the trip as the ball travels upwards, it starts at a velocity of 25 m/s, and at the peak of its trajectory, it will have a velocity of 0 m/s. We can use the formula vf = vi + at to calculate time. Plugging in these values and the acceleration due to gravity yields: vf = vi + at 0 m/s = 25 m/s + (-10 m/s2) t -25 = -10t 2.5 seconds = t However, 2.5 seconds is only the time needed to reach the top of the trajectory. We must multiply this number by 2 to account for the time it takes the ball to come back down, yielding 5 seconds.
If a bullet is fired at 10 m/s at an angle of 60 degrees above the horizontal, what will be its horizontal velocity right before it hits the ground? (Note that sin 60 = 0.87 and cos 60 = 0.5.)
Because there are no forces exerted on this object in the horizontal direction, the initial horizontal velocity will be the same as the final horizontal velocity. Therefore, we just need to break down the initial velocity given into its components. We can calculate the x-component as follows: vix = vi cosθ vix = (10 m/s) cos (60) vix = (10 m/s) (0.5) vix = 5 m/s
If a baseball player throws a ball at 10 m/s at 60 degrees above the horizontal, what is the horizontal distance that the ball will travel before it lands? (sin 60 = 0.87 and cos 60 = 0.5)
In order to answer this question, we need to find: (1) the initial velocity of the ball in the y-direction, (2) the total time the ball spends in the air, and (3) the velocity of the ball in the x-direction. First, let’s calculate the ball’s initial vertical velocity: viy = (vi)sinθ viy = (10 m/s) sin (60) viy = (10 m/s) (0.87) viy = 8.7 m/s Now, we can calculate the total time that the ball spends in the air. We know that the initial vertical velocity is 8.7 m/s, and at the peak of the flight, the vertical velocity will be 0 m/s. The acceleration will remain constant at -10 m/s2. We can plug these values into the equation vf = vi + at: vf = vi + at (0 m/s) = (8.7 m/s) + (-10 m/s2) t -8.7 = -10t t = 0.87 seconds. Remember, this is the time required for only the first half of the flight. The total time in the air will be double this value, or 1.74 seconds. Next we can calculate the ball’s velocity in the x-direction, which will remain constant: vx = vicosθ vx = (10 m/s) cos (60) vx = (10 m/s) (0.5) vx ≈ 5 m/s Finally, we can combine this information to solve for the distance that the ball travels. We know that the horizontal velocity is about 5 m/s and that the ball is in the air for 1.74 seconds. We can thus calculate: d= vt d = (5 m/s)(1.74 s) d = 8.7 meters
If I fire a bullet to my right and drop a bullet to my left at the same time, which will hit the ground first?
They will hit at the same time
What is terminal velocity? `
Point where Fg=Fairresistance so there is no net force acting on the object so acceleration is 0. It still has velocity, just no acceleration
In the graph, what would the area under the curve tell you?
The units of the area under the graph would be y*x units so it would be distance*time m*s. It would ultimately give you distance.
