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Lecture 04:. Kinematics + Dynamics. Kinematics Equations constant acceleration Dynamics Newton’s Second Law Non-zero acceleration. Equations for Constant Acceleration. x = x 0 + v 0 t + ½ at 2 v = v 0 + at v 2 = v 0 2 + 2a(x-x 0 ). x is final position x o is initial position

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## Kinematics + Dynamics

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**Lecture 04:**Kinematics + Dynamics • Kinematics Equations • constant acceleration • Dynamics • Newton’s Second Law • Non-zero acceleration**Equations for Constant Acceleration**• x = x0 + v0t + ½ at2 • v = v0 + at • v2 = v02 + 2a(x-x0) • x is final position • xo is initial position • v is final velocity • v0 is initial velocity • a is acceleration • t is time**Kinematics Example**• A car is traveling 30 m/s and applies its breaks to stop. Assuming constant acceleration of -6 m/s2, how long does it take for the car to stop, and how far does it travel before stopping? • x = x0 + v0t + ½ at2 • v = v0 + at • v2 = v02 + 2a(x-x0) • Begin by using the second equation to find the time: v = v0 + a t (0 m/s) = (30 m/s) + (-6 m/s2) t t = 5 s • Then use the first equation to find the distance: x = x0 + v0t + ½ a t2 x = (0 m) + (30 m/s)(5s) + ½ (-6 m/s2) (5 s)2 x = 75 m**Dynamics: F = ma**• We have already dealt with situations where a = 0. • But when the net force is not zero, there IS an acceleration!**FT**y Fg x Dynamics Example • A tractor is pulling a trailer with a constant acceleration. If the forward acceleration is 1.5 m/s2, Calculate the force on the trailer (m = 400 kg) due to the tractor. (Consider just the trailer.) FBD: x-direction: FN F = ma FT = ma FT = 600 N F = ma FN- Fg = 0 FN = 3920 N FT = 600 N y-direction:**Summary of Concepts**• Constant Acceleration • x = x0 + v0t + ½ at2 • v = v0 + at • v2 = v02 + 2a(x-x0) • F = m a • Draw Free Body Diagram • Write down equations • Solve**Kinematics Example**• A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red?) Note: even after accelerating, the car is still traveling safely under the speed limit… • There are two ways to solve this…I will use one method and you will use the other!**Kinematics Example**• A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red?) Note: even after accelerating, the car is still traveling safely under the speed limit… • I will do this in two steps: • First, using v2 = v02 + 2 a (x-x0), we can solve for the final velocity. • Second, using v = v0 + a t, we can solve for the time.**Kinematics Example**• A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red?) Note: even after accelerating, the car is still traveling safely under the speed limit… v2 = v02 + 2 a (x-x0) v2 = (15 m/s)2 + 2 (4 m/s2) (18 m – 0 m) v = 19.2 m/s v = v0 + a t (19.2 m/s) = (15 m/s) + (4 m/s2) t t = 1.05 s The car does not make it before the light turns red… It should have just stopped!

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