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Lecture 4: More kinematics. Position vector points from the origin to a location. The displacement vector points from the original position to the final position. Displacement and change in displacement. Average velocity vector: So is in the same direction as . t 1. t 2.
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Lecture 4: More kinematics
Position vector points from the origin to a location. The displacement vector points from the original position to the final position. Displacement and change in displacement
Average velocity vector: So is in the same direction as . t1 t2 Average Velocity
Instantaneous velocity vector v is always tangent to the path. t1 t2 Instantaneous
Average Acceleration Average acceleration vector is in the direction of the change in velocity:
Relative Motion Velocity vectors can add, just like displacement vectors The speed of the passenger with respect to the ground depends on the relative directions of the passenger’s and train’s speeds:
Relative Motion This also works in two dimensions:
You are riding on a Jet Ski at an angle of 35° upstream on a river flowing with a speed of 2.8 m/s. If your velocity relative to the ground is 9.5 m/s at an angle of 20.0° upstream, what is the speed of the Jet Ski relative to the water? (Note: Angles are measured relative to the x axis shown.)
Now suppose the Jet Ski is moving at a speed of 12 m/s relative to the water. (a) At what angle must you point the Jet Ski if your velocity relative to the ground is to be perpendicular to the shore of the river? (b) If you increase the speed of the Jet Ski relative to the water, does the angle in part (a) increase, decrease, or stay the same? Explain. (Note: Angles are measured relative to the x axis shown.)
2-Dimensional Motion (sections 4.1-4.5)
Vector Components II a) 30° b) 180° c) 90° d) 60° e) 45° A certain vector has x and y components that are equal in magnitude. Which of the following is a possible angle for this vector in a standard x-y coordinate system?
Vector Components II a) 30° b) 180° c) 90° d) 60° e) 45° A certain vector has x and y components that are equal in magnitude. Which of the following is a possible angle for this vector in a standard x-y coordinate system? The angle of the vector is given by tan Θ = y/x. Thus, tan Θ = 1 in this case if x and y are equal, which means that the angle must be 45°.
Acceleration and Velocity Vectors • a) point 1 • b) point 2 • c) point 3 • d) point 4 • e) I cannot tell from that graph. Below is plotted the trajectory of a particle in two dimensions, along with instantaneous velocity and acceleration vectors at 4 points. For which point is the particle speeding up?
Acceleration and Velocity Vectors • a) point 1 • b) point 2 • c) point 3 • d) point 4 • e) I cannot tell from that graph. Below is plotted the trajectory of a particle in two dimensions, along with instantaneous velocity and acceleration vectors at 4 points. For which point is the particle speeding up? At point 4, the acceleration and velocity point in the same direction, so the particle is speeding up
a) b) c) d) Fueling Up The tanker and fighter jet are both travelling with a level velocity of 150 m/s. Which choice would best approximate the velocity of the fuel in the nozzle?
a) b) c) d) Fueling Up The tanker and fighter jet are both travelling with a level velocity of 150 m/s. Which choice would best approximate the velocity of the fuel in the nozzle?
The Components of Velocity Vector v vy vx Motion along each direction becomes a 1-D problem
Projectile Motion: objects moving under gravity y g • Assumptions: • ignore air resistance • g = 9.81 m/s2, downward • ignore Earth’s rotation • y-axis points upward, x-axis points horizontally • acceleration in x-direction is zero • Acceleration in y-direction is -9.81 m/s2 vy x vx
Relativity Car a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart e) it remains at rest A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball?
when viewed fromtrain when viewed fromground Relativity Car a) it depends on how fast the cart is moving b) it falls behind the cart c) it falls in front of the cart d) it falls right back into the cart e) it remains at rest A small cart is rolling at constant velocity on a flat track. It fires a ball straight up into the air as it moves. After it is fired, what happens to the ball? In the frame of reference of the cart, the ball only has a vertical component of velocity. So it goes up and comes back down. To a ground observer, both the cart and the ball have the same horizontal velocity, so the ball still returns into the cart.
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Launch angle: direction of initial velocity with respect to horizontal
Zero Launch Angle In this case, the initial velocity in the y-direction is zero. Here are the equations of motion, with x0 = 0 and y0 = h:
Zero Launch Angle Eliminating t and solving for y as a function of x: This has the form y = a + bx2, which is the equation of a parabola. The landing point can be found by setting y = 0 and solving for x:
Trajectory of a zero launch-angle projectile horizontal points equally spaced vertical points not equally spaced parabolic y = a + bx2
Which ball will reach the end first? ball 1 ball 2 Drop and not a) Top ball (Ball 1) b) Bottom ball (Ball 2) c) They will arrive at the same time d) impossible to say by the given information
y vx vy=0 vx x vy=0
vx y Δx vy=0 vx v vy x vy=0
y v vy=0 x vy=0
Monkey and the Hunter a) 15 m/s b) 24 m/s c) 38 m/s d) 47 m/s e) 73 m/s If the hunter is 12 meters from the target, the gun is inclined at 10o to the horizontal, and the target drops 50.0 cm before being struck, what is the muzzle velocity of the hunter’s gun?
Monkey and the Hunter a) 15 m/s b) 24 m/s c) 38 m/s d) 47 m/s e) 73 m/s If the hunter is 12 meters from the target, the gun is inclined at 10o to the horizontal, and the target drops 50.0 cm before being struck, what is the muzzle velocity of the hunter’s gun? • The time it takes to fall 50 cm, starting with v0y = 0, is given by • 0.50 m = g/2 t2 • t = 0.319 s • So the (constant) horizontal velocity of the projectile must be • vx = 12 m / 0.319 s = 37.6 m/s • And the total initial velocity can be found from the launch angle: • v0 = vx / cos(10o) = 38 m/s v0 v0y 10o vx
-g v0Sin(θ) v0Cos(θ) General Launch Angle In general, v0x = v0 cos θ and v0y = v0 sin θ This gives the equations of motion:
Range: the horizontal distance a projectile travels As before, use (y = 0 at landing) and Eliminate t and solve for x when y=0
Sin(2) θ Range is maximum at 45o
Range Gun a) 10 degrees b) 22.5 degrees c) 60 degrees d) 75 degrees e) one would also need the launch velocity of the range gun to know If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters?
Range Gun a) 10 degrees b) 22.5 degrees c) 60 degrees d) 75 degrees e) one would also need the launch velocity of the range gun to know If the range gun launches a ball 6 meters with a launch angle of 45 degrees, at which of these angles should a ball be launched to land in a bucket at 3 meters? The range is proportional to sin2θ, so to travel half the distance, the ball would need to be launched with sin2θ = 0.5. sin(75o) = 0.5, so θ=75o Note: there are two angles that would work [sin(30o) = 0.5 also]. How are the two solutions different?
Dropping the Ball III a) just after it is launched b) at the highest point in its flight c) just before it hits the ground d) halfway between the ground and the highest point e) speed is always constant A projectile is launched from the ground at an angle of 30°. At what point in its trajectory does this projectile have the least speed?
Dropping the Ball III a) just after it is launched b) at the highest point in its flight c) just before it hits the ground d) halfway between the ground and the highest point e) speed is always constant A projectile is launched from the ground at an angle of 30º. At what point in its trajectory does this projectile have the least speed? The speed is smallest at the highest point of its flight path because they-component of thevelocity is zero.
h a b c d) all have the same hang time Punts I Which of the three punts has the longest hangtime?
Follow-up: Which one had the greater initial velocity? h a b c d) all have the same hang time Punts I Which of thethree puntshas thelongest hangtime? The time in the air is determined by the vertical motion! Because all of the punts reach the same height, they all stay in the air for the same time.
•• On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25 m/s at an angle of 35.0° above the horizontal. If she is in flight for 0.616 s, how high above the water was she when she let go of the rope?
Time to hit the water: t=0.616 Initial velocity: 2.25 m/s at 35o above the horizontal y = (v0 sinθ) t - 1/2 g t2 y = 2.25 m/s * sin(35o) * (0.616 s) - 1/2 (9.8 m/s2) (0.616 s)2 = - 1.07 m At time t= 0.616 s, the girl is 1.07 m below her starting position, so her initial position was 1.07m above the water.