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## Homework #5 – due Friday , 10/7 Reading assignment: 3.1-3.3.5

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**Homework #5 – due Friday, 10/7**Reading assignment: 3.1-3.3.5 Questions: 3.2, 4, 12, 16, 18, 19, 27, 36, 39, 82 – the solutions are on the school website. Homework – due Tuesday, 10/11 – 11:00 pm Mastering physics wk 5**Chapter 3**Position, velocity, acceleration vectors Projectile motion Relative motion**Homework #5 – due Friday, 10/7**Reading assignment: • 3.1-3.3, 3.5 Questions: 3.2, 4, 12, 16, 18, 19, 27, 36, 39, 82 – the solutions are on the school website. • Homework – due Tuesday, 10/11 – 11:00 pm • Mastering physics wk 5**(2D)**Direction of v is tangent to path of particle at P**Test your understanding**• In which of these situations would the average velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval: • A body moving along a curved path at constant speed • A body moving along a curved path and speeding up • A body moving along a straight line at constant speed • A body moving along a straight line and speeding up.**2b**3c t = example If r = bt2i+ ct3j Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes? v = dr/dt = 2bt i + 3ct2 j = 1**Example 3.2**• Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s. • Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.**The effect of acceleration’s directions**• When acceleration is perpendicular to particle’s velocity: velocity’s magnitude does not change, only its direction changes, particle moves in a curved path at constant speed. • When acceleration is parallel to particle’s velocity: velocity’s magnitude changes only, its direction remains the same, particle moves in a straight line with changing speed.**decreasing speed**Increasing speed Constant speed**Example 3.3**• Given: • Find the parallel and perpendicular components of the instantaneous acceleration at t = 2.0 s**Test your understanding 3.2**• A sled travels over the crest of a snow-covered hill. The sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)**3.3 projectile motion**• A projectile is any body that is given an initial velocity and then follows a path determined entirely by the effects of gravitational acceleration and air resistance. • Example: a batted baseball, a thrown football, a package dropped from an airplane, a bullet shot from a rifle. • The path of a projectile is called a trajectory.**x (t) = yo + voxt**y (t) = yo + voyt – ½ gt2 z (t) = 0**Example 3.5**Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance. The acceleration at points G, H, I are the same: ax = 0; ay = -9.8 m/s2**Example 3.6:a body projected horizontally**• A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position,distance and velocity from the edge of the cliff, and velocity after 0.50 s.**r = √4.52 + (-1.2 m)2**• Position: α0 = 0o; vo = 9.0 m/s; voy = 0; t = 0.5 s x = vocosα0 ∙t = (9.0 m/s)cos0o(0.5 s) = 4.5 m y = vosinα0 ∙t - ½ gt2= - ½ (9.8 m/s2)(0.5 s)2 = -1.2 m • distance: r = 4.7 m**v = √(9.0m/s)2 + (-4.9 m/s)2**• velocity: α0 = 0o; vo = 9.0 m/s; t = 0.5 s vx = vo = 9.0 m/s vy = - gt= - (9.8 m/s2)(0.5 s)= - 4.9 m/s • magnitude: v = 10.2 m/s • direction: α = 29o below horizontal**Example: 3.7 – height and range of a projectile I – a**batted baseball**a.**• Position: α0 = 53.1o; vo = 37.0 m/s; t = 2.00 s x = (37.0 m/s)(cos53.1o)(2.00 s) = 44.4 m y = (37.0 m/s)(sin53.1o)(2.00 s) - ½ (9.80 m/s2)(2.00 s)2 = 39.6 m • velocity: vx = (37.0 m/s)(cos53.1o) = 22.2 m/s vy = (37.0 m/s)(sin53.1o) - (9.80 m/s2)(2.00 s)= 9.99 m/s v = 24.3 m/s α = 24.2o above horizontal**b. At highest point, vy = 0;**tmax = (37.0 m/s)(sin53.1o)/(9.80 m/s2) = 3.02 s The height reached at this point: ymax = ½ [(37.0 m/s)(sin53.1o)] 2 / (9.80 m/s2) = 44.7 m**c. Range**xmax = 134 m**You toss a ball from your window 8.0 m above the ground.**When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.**Given: y = - 8.0 m; α0 = - 20o; v0 = 10.0 m/s;**Unknown: x = ? To find t, we use -8.0 m = (10.0 m/s)sin(-20o)t – ½ (9.8 m/s2)t2 t1 = -1.7 s; t2 = 0.98 s x = (10.0 m/s)cos(-20o)(0.98 s) = 9.2 m