Velocity-time graphs For a vel-t graph with: Constant acceleration, a.

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Presentation transcript:

Velocity-time graphs For a vel-t graph with: Constant acceleration, a. initial velocity, vi. final velocity, vf. And a graph that looks like the one on the board.

Velocity-time graphs From the slope-intercept equation for a line: y = mx + b. y = velocity (v) x = time (t) m = acceleration (a) b = initial velocity (vi) Equation becomes: v = vi + at Solving for the final velocity (vf) at any time t, this equation is rewritten as vf = vi + at.

Kinematics vf = vi + at is the first kinematic equation. Kinematic equations describe the motion of an object undergoing constant (unchanging) acceleration. There are 4 kinematic equations. Let’s derive the rest of them!

Kinematics On a vel-t graph, the displacement (distance) is found by calculating the area under the curve. This area is made up of a triangle (A) and a rectangle (A ). What is the area of the triangle? A = ½ (vf – vi)t What is the area of the rectangle? A = vit

Kinematics Total area = A + A = ? Distance = A + A = ½ (vf – vi)t + vit d = ½ (vf – vi)t + vit Solve for d. d = ½ (vi + vf)t This is the 2nd kinematic equation.

Kinematics Substitute vf=vi + at into d=½ (vi + vf)t d=½ (vi + vi + at)t d=½ (2vi+ at)t d=vi t+ ½ at2 This is the 3rd kinematic equation

Kinematics Solve vf = vi + at for t. t = (vf - vi)/a Substitute into d=½ (vi + vf)t d=½ (vi + vf) (vf - vi)/a 2ad=vf2 - vi2 Rewritten as vf2 = vi2 + 2ad This is the 4th kinematic equation.

Kinematics Table to put into notes To solve equations: Write down what you know Write down what you are seeking Write down the equation that contains those variables.

Kinematics Examples Aircraft carrier overview Launching area, landing area, blast shields

Kinematics Examples - Nimitz Runway length is 310 ft or 94.5 m. Initial velocity = 0 m/s Final velocity = 170 mph or 76.0 m/s A steam-powered catapult accelerates the planes to launch by attaching to the nose cone.

Kinematics Examples - Nimitz What is the acceleration required for a successful launch? You know: vi = 0 m/s, vf = 76.0 m/s, and d = 94.5 m. You are seeking a. What equation contains those variables? vf 2 = vi 2 + 2ad (76.0 m/s)2 =(0 m/s)2 + 2a(94.5 m) a = 30.6 m/s2 (with 3 sig figs) The force felt by the pilot is about 3 g’s.

Kinematics Examples - Nimitz What is the time during which this acceleration takes place? You know: vi = 0 m/s, vf = 76.0 m/s, and d = 94.5 m. You are seeking t. What equation contains those variables? d = ½ (vi + vf)t 94.5 m = ½ (0 m/s + 76.0 m/s)t t = 2.49 s (with 3 sig figs) Yes – that IS 0 to 170 mph in 2.5 seconds!

Kinematics Examples - Nimitz Aircraft land by catching a tailhook on an arresting wire. The landing runway is 315 ft or 96 m. The initial velocity is 150 mph or 67 m/s. The total time to land is 2.0 s.

Kinematics Examples - Nimitz What was the acceleration during this time? You know: vi = 67 m/s, d = 96 m, and t = 2.0 s. You are seeking a. What equation contains those variables? d=vi t+ ½ at2 96 m = (67 m/s)(2.0 s) + ½ a(2.0 s)2 a = -19 m/s2 (with 2 sig figs) Note the negative acceleration because the aircraft is actually slowing down! The force felt by the pilot is about 2g’s.

Kinematics Examples A car initially moving at 17 m/s accelerates at 12 m/s2 for 180 m. During what time did this acceleration take place? You know: vi = 17 m/s, a = 12 m/s2, and d = 180 m. You are seeking t. What equation contains those variables? d=vi t+ ½ at2 180 m = (17 m/s) t + ½ (12 m/s2) t2 How will you solve this equation for t? Use the quadratic equation. How to do on TI calcs t = 4.2 s (with 2 sig figs)

References for images http://www.aviationspectator.com/files/images/E-2C-Hawkeye-108.preview.jpg http://farm3.static.flickr.com/2545/3841972788_7e83d5b964_o.jpg US Dept of Defense