BIOM 121 1nd Edition Lecture 1 Outline of Current Lecture I. Kinematics in 1 Dimensiona. Kinematicsb. Dynamicsc. Motiond. Displacemente. Average Velocityf. Instantaneous Velocityg. Accelerationh. Position for Constant Accelerationi. Constant AccelerationII. Example ProblemCurrent LectureI. Kinematics in 1 Dimensiona. Kinematics: subject describing motionb. Dynamics: cause of motion c. Motion is described in terms of a coordinate system that defines position xd. Displacement: change in distance from ending point to starting pointThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.∆ x=xf−xie. Average Velocity: is the displacement over a period of timevavg=∆ x∆ tf. Instantaneous Velocity: speedg. Acceleration: rate of change of velocityv =v0+ath. Position for constant acceleration∆ x=v0t +(12)a t2v2=v02+2 a(∆ x)i. Constant accelerationi. Distance vs. time graph would have an exponential graphii. Velocity vs. time graph would have a linear increasing graphiii. Acceleration vs. time graph would be a horizontal line graphII. Example:a. A bicyclist is traveling towards a gate at 10 m/s and is 100 m from the gate. A dog is also traveling towards the gate at 12 m/s and is 110 m from the gate. Where does the dog catch up with the bicyclist? (Calculate both time and distance)i. Write down what we know.1. a = 0, vd = 12 m/s, vb = 10 m/s, xd = 0 m, xb = 10 m, t = ?ii. Write an equation to solve for the variables.1.∆ x=v0t +(12)a t22.xf−xi=v0t+(12)(0)t23.xf=xi+v04. Position for Dog: x = 12t5. Position for Bicyclist: x = 10 + 10tiii. Solve the equations1. 12t = 10 + 10t2. t = 5 seconds, x = 60
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