 Lecture 4: More kinematics. Displacement and change in displacement Position vector points from the origin to a location. The displacement vector points.

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Lecture 4: More kinematics

Displacement and change in displacement Position vector points from the origin to a location. The displacement vector points from the original position to the final position.

Average Velocity t1t1 t2t2 Average velocity vector: So is in the same direction as.

Instantaneous velocity vector v is always tangent to the path. Instantaneous t1t1 t2t2

Average Acceleration Average acceleration vector is in the direction of the change in velocity:

Relative Motion The speed of the passenger with respect to the ground depends on the relative directions of the passenger’s and train’s speeds: Velocity vectors can add, just like displacement vectors

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 )

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? a) 30° b) 180° c) 90° d) 60° e) 45° Vector Components II Vector Components II

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? a) 30° b) 180° c) 90° d) 60° e) 45° 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°. Vector Components II Vector Components II

a) point 1 b) point 2 c) point 3 d) point 4 e) I cannot tell from that graph. Acceleration and Velocity Vectors 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?

a) point 1 b) point 2 c) point 3 d) point 4 e) I cannot tell from that graph. Acceleration and Velocity Vectors 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

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?a) b)c)d)

The Components of Velocity Vector Motion along each direction becomes a 1-D problem vxvx vyvy v

Projectile Motion: objects moving under gravity Assumptions: ignore air resistance g = 9.81 m/s 2, 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/s 2 x y g vyvy vxvx

Relativity Car 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? 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? 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 when viewed from train when viewed from ground 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. Relativity Car

These, then, are the basic equations of projectile motion:

+ =

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 x 0 = 0 and y 0 = h:

Zero Launch Angle Eliminating t and solving for y as a function of x: This has the form y = a + bx 2, 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 + bx 2

ball 1 ball 2 Which ball will reach the end first? 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 Drop and not

x y vxvx vxvx v y =0

x y ΔxΔx vxvx vxvx vyvy v

x y v

General Launch Angle

Monkey and the Hunter If the hunter is 12 meters from the target, the gun is inclined at 10 o to the horizontal, and the target drops 50.0 cm before being struck, what is the muzzle velocity of the hunter’s gun? a) 15 m/s b) 24 m/s c) 38 m/s d) 47 m/s e) 73 m/s

Monkey and the Hunter If the hunter is 12 meters from the target, the gun is inclined at 10 o to the horizontal, and the target drops 50.0 cm before being struck, what is the muzzle velocity of the hunter’s gun? a) 15 m/s b) 24 m/s c) 38 m/s d) 47 m/s e) 73 m/s The time it takes to fall 50 cm, starting with v 0y = 0, is given by 0.50 m = g/2 t 2 t = 0.319 s So the (constant) horizontal velocity of the projectile must be v x = 12 m / 0.319 s = 37.6 m/s And the total initial velocity can be found from the launch angle: v0 = v x / cos(10 o ) = 38 m/s vxvx v 0y v0v0 10 o

-g v 0 Sin(θ) v 0 Cos(θ) General Launch Angle In general, v 0x = v 0 cos θ and v 0y = v 0 sin θ This gives the equations of motion:

Range: the horizontal distance a projectile travels As before, use and Eliminate t and solve for x when y=0 (y = 0 at landing)

θ Sin  Range is maximum at 45 o

Range Gun 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? 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

Range Gun 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? 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 The range is proportional to sin2θ, so to travel half the distance, the ball would need to be launched with sin2θ = 0.5. sin(75 o ) = 0.5, so θ=75 o Note: there are two angles that would work [sin(30 o ) = 0.5 also]. How are the two solutions different?

Symmetry in projectile motion

Dropping the Ball III 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? 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 º. At what point in its trajectory does this projectile have the least speed? 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 The speed is smallest at the highest point of its flight path because the y-component of the velocity is zero. Dropping the Ball III

? Which of the three punts has the longest hang time? Punts I d) all have the same hang time abc h

Which of the three punts has the longest hang time? d) all have the same hang time abc h 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. Follow-up: Which one had the greater initial velocity? Punts I

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?

y = (v 0 sinθ) t - 1/2 g t 2 y = 2.25 m/s * sin(35 o ) * (0.616 s) - 1/2 (9.8 m/s 2 ) (0.616 s) 2 = - 1.07 m Time to hit the water: t=0.616 Initial velocity: 2.25 m/s at 35 o above the horizontal 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.

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall?

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14 m/s at an angle of 34° above the horizontal. (a) How long does it take for the ball to reach the wall if it is 3.8 m away? (b) How high is the ball when it hits the wall? d = 3.8 m v 0 = 14 m/s θ = 34 o v 0x = (14 m/s) cos(34 o ) = 11.6 m/s d = v 0x t t = d/v 0x = (3.8m) / (11.6 m/s) = 0.33 s v 0y = (14 m/s) sin(34 o ) = 7.83 m/s h = v 0y t - 1/2 g t 2 = (7.83 m/s) (0.33s) -1/2 (9.8 m/s 2 ) (0.33 s) 2 = 2.0 m

In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall (3.8 m away) with a speed of 14 m/s at an angle of 34° above the horizontal. (a) what are the magnitude and direction of the ball's velocity when it strikes the wall? (b) Has the ball reached the highest point of its trajectory at this time? Explain. d = 3.8 mv 0 = 14 m/sθ = 34 o v 0x = (14 m/s) cos(34 o ) = 11.6 m/s v 0y = (14 m/s) sin(34 o ) = 7.83 m/s t = d/v 0x = (3.8m) / (11.6 m/s) = 0.33 s v y = v 0y - g t = 7.83 m/s - (9.8m/s 2 ) (0.33s) = 4.6 m/s v x = v 0x The vertical component of velocity is still positive, that is, the ball is still going up. So the ball has not yet reached its highest point.

Equations for Test 1

 Assignment 2 on MasteringPhysics due Monday, September 6 (12:59 pm).  Happy Labor Day!  Exit using the rear doors!

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