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## Physics

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**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Speed …that’s how fast I am going… Velocity …that’s my speed and direction… SCALAR VECTOR**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**How can we illustrate motion graphically? Position vs Time graphs _Very useful for rectilinear motion for they show the way in which motion took place (continuous or not, slow, fast, constant or not, etc) _Position (d) is the y-axis, whereas time (t) is the x-axis _Slope gives you speed**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Description of motion ( Position vs time graph): _60 m in 10 sec _stops (0 m) for 5 sec _100 m in 25 sec (opposite direction = negative direction of axis) _40 m in 15 sec Is it possible to have a vertical section in this graph? Why?**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Description of motion ( Position vs time graph): _ _ _ 12 m in 4 sec 0 m in 4 sec 12 m in 12 sec (opposite direction)**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.10 – 3.11, Ex 3.2 (b, d) (use “a” as an example to draw the trajectory)**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**What is average speed? Average speed = Mathematically: : distance between initial and final position : time required to travel said distance : average speed**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Speed and unit conversion _Car travelling at 50 m/s, how much is the speed in km/h? = = 180 km / h**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Examples of calculations (p. 3.19): At an F1 race held on a 4.85 km circuit JV’s car completes 50th lap in 1 h 12 min 15 s. Knowing that the previous one ended at 1h 10 min 45 s, calculate his average speed for the 50th lap. = t2 - t1 = (135s – 45s) = 90s (90s / 3600s = 0.025h)**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.19, Ex 3.9**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Speed (scalar) Ratio of total distance and time interval Right or Wrong? _v = 3km/h _v = 7 m/s, 20° Velocity (vector) Ratio of total displacement and time interval Right or Wrong? _ = 3km/h _ = 7 m/s, 20°**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Parallel vectors An airport moving sidewalk moves with velocity of 4 km/h with respect to the floor. A man walks with velocity of 3 km/h with respect to moving sidewalk. What’s his velocity with respect to the floor? Vt = Vs + Vm (vector symbol) Vt = 7 km/h Opposite vectors An airport moving sidewalk moves with velocity of 4 km/h with respect to the floor. A child walks with velocity of 6 km/h with respect to moving sidewalk, in opposite direction to it. What’s his velocity with respect to the floor? Vt = Vc - Vs(vector symbol) Vt = 2 km/h**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.30, Ex 3.18 (info on 3.17)**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Uniform rectilinear motion: constant velocity (constant speed and direction) _Simplest motion _Limited duration _Position vs Time graphs will provide info on speed, therefore velocity. _Positive slopes, positive axis direction (vice verse) _Slope = velocity**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**P V V P t t t t**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.45, Ex 3.29 Page 3.47, Ex 3.31 Page 3.49, Ex 3.32**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**We can find velocity using a P vs t graph for uniform rectilinear motion. But, can we find position using a V vs t graph for uniform rectilinear motion? YES! Calculating the “area under the curve” for V vs t graphs _See Fig. 3.14 a and 3.14b, p. 3.50 _3.14a (P vs t) train travelled 2m in 5 s _3.14b (V vs t) at 5s, perpendicular line to cut graph. _3.14b Rectangle (or square) formed _Calculating area of ABCD implies V * t (m/s * s), so we obtained position (in meters) _For 5 s (V vs t), we get d = (5s * 0.4m/s) d = 2m (check 3.14a!) _Try for 3s now!**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**In uniform rectilinear motion, the area under the curve of a V vs t graph provides not only the value of distance (at any given time), but also displacement since distance and displacement are equal for this type of motion *Displacement and distance travelled are both equal and calculated as the area under the curve for any type of rectilinear movement (no change in direction)**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**When direction changes: _portion of area under the curve on positive side of axis (positive value) _portion of area under the curve on negative side of axis (negative value) _distance travelled is then the sum of the absolute values of either area under the curve _displacement is the algebraic sum of said areas under the curve**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**_0 to 3s, vΔt = 1m/s * 3s = 3m _3 to 6s, vΔt = 2m/s * 3s = 6m _6 to 8s, vΔt = 1m/s * 2s = 2m _8 to 9s, vΔt = 0m/s * 1s = 0m _9 to 10s, vΔt = -2m/s * 1s = -2m _10 to 11s, vΔt = -3m/s * 1s = -3m _11 to 15s, vΔt = -4m/s * 4s = -16m Distance travelled: 3m+6m+2m+(2m)+(3m)+(16m) = 32m Displacement: (3+6+2)m – (2+3+16)m = -10m**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.54, Ex 3.34**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Motion Equation: v = Δd / Δt In many cases t1 = 0, so d2= v t2 + d1 Δd = v * Δt or just: d = vt+ d1 d2 – d1= v * Δt d2= vΔt + d1**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**A cyclist travels at 12 m/s in uniform rectilinear motion. We begin observing his motion when he is 250m from start point. • Motion equation? • Position of cyclist at 10s of observation • When will he be 1000m from start line? d = 12t + 250 d = 12(10) + 250 d = 370m d = 12t + 250 t = (d – 250) / 12 t = 62.5s**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice Exercises: Page 3.57, Ex 3.35**PHS 5042-2 Kinematics & MomentumSpeed & Velocity**Practice exercises: Page 3.100 – 3.104, Ex 3.52 – 3.56 and 3.58