Kinematics - 4 - Acceleration

Accelerated Motion (continued) 3.5 Average vs Instantaneous Velocity Recall:   Acceleration is the rate at which velocity increases or decreases If an object is accelerating it is not experiencing uniform motion Average Velocity (for uniform or zero acceleration): v (m/s) v (m/s) t (s) - - t (s) Area under curve = displacement Left rectangle (uniform motion) right triangle (non-uniform motion) can be equal if vleft = average(vright ) v (m/s) v (m/s) t (s) t (s) d = vavgt ( v1 + v2 ) vavg = 2 ( v1 + v2 )t d = 2 - Average velocity only requires total displacement and total time elapsed. Franklin runs 5.0 km [E] from his house to the library at 8.0 km/h. He then runs halfway back at 6.0 km/h. Determine his average velocity. The key to average velocity problems is TOTAL time and TOTAL displacement. Out: 1. Find time for each leg: Δd Δd v = 8.0 km/h Δt = Δt = v v Δd = 5.0 km 2.5 5.0 = = Δt = ? 6.0 8.0 = 0.63 h = 0.42 h Back: v = 6.0 km/h 2. Total time = 0.63 + 0.42 = 1.05 h Δd = 2.5 km 3. Find average velocity: v = Δd Δt Δt = ? total avg total 5.0 - 2.5 1.05 = 2.4 km/h [E] = Creating d vs t from v vs t graphs - Create table of values of displacement from v vs t graph (areas), plot as d vs t t (s) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 d (m) Instantaneous Velocity - Linear non-zero slope velocity graph means a quadratic displacement graph! - How to find velocity of curved displacement graph (“slope” of a curve?) - Find tangent line at desired time. 8 Instantaneous Velocity Position vs time Find the velocity at 4.0 s. 100 90 Positiion (m [E]) Slope of tangent line of position vs time graph gives the instantaneous velocity! Δy a= Δx (62 - 0) m = (10 - 2) s = 7.8 m/s 80 70 60 50 40 30 Or 7.8 m/s[E] 20 10 0 0 2 4 6 8 10 Time (s) 9 3.6: Equations for uniformly accelerated motion  To solve acceleration questions we can use these equations ( v1 + v2 )t d = v2 = v1 + at 2 to derive these equations: 1 d = v1t + a (t )2 2 (v2 ) 2 = (v1 ) + 2ad 2 ( v1 + v2 )t d = Substitute v2 = v1 + at to get into 2 1 d = v1t + a (t )2 2 Substitute v2 − v1 t = a to get (v2 ) 2 ( v1 + v2 )t d = into 2 = (v1 ) + 2ad 2 Ex #1 A car acquires a velocity of 32 m/s by accelerating at 4.0 m/s2 for 5.0 s. What was its initial velocity? v2 = 32m/s a = 4.0m/s2 Δt = 5.0s v1 = ? v2 = v1 + at 32 = v1 + (4)(5) v1 = 12m / s Ex #2 v1 = 36m/s Δ t = 3.0s v2 = 0m/s Δd = ? a=? The brakes are applied on a car traveling at 36.0 m/s. The car stops in 3.0 s. What is the car’s displacement & acceleration? v2 = v1 + at 0 = 36 + a (3) a = −12m / s 2 1 2 d = v1t + a (t ) 2 1 2 d = (36)(3) + ( −12)(3) 2 d = 54m Ex #3 How long does it take an airplane accelerating from rest at 5.0 m/s2 to travel a distance of 360.0 m? a= 5m/s2 v1 = 0m/s Δd = 360m Δt = ? 1 2 d = v1t + a (t ) 2 1 2 360 = (0)t + (5)(t ) 2 t = 12s Ex #4 A person biking at a velocity of 10.0 m/s speeds up for a distance of 24.0 m, accelerating at a rate of 3.00 m/s2. What is the bike’s final velocity? v1 = 10m/s Δd = 24m a = 3m/s2 v2 = ? (v2 ) = (v1 ) + 2ad 2 2 (v2 ) = (10) + 2(3)(24) 2 2 v2 = 15.6m / s Ex #5 − b  b 2 − 4ac 2a How long does it take a car traveling at 22.0 m/s accelerating at 4.0 m/s2 to travel a distance of 20.0 m? v1 = 22m/s a = 4m/s2 Δd = 20m Δt = ? 1 2 d = v1t + a (t ) 2 1 2 20 = 22t + (4)(t ) 2 0 = 2t + 22t − 20 t = 0.84s or − 11.84s 2 Ex #6 A driver notices an obstacle in the road. Her reaction time is 0.700 s. While braking, her car slows at a rate of 7.50 m/s2. If her initial velocity was 50.0 km/h, what is the distance traveled from the time she notices the obstacle until the car stops? v1 = 13.89m/s v2 = 0m/s a = -7.50m/s2 Δd = ? v1 = 13.89m/s Δt = 0.700s Δd = ? (v2 ) = (v1 ) 2 2 + 2ad (0) = (13.89) + 2(−7.5)(d ) d = 12.9m 2 d v= t d 13.89 = 0.700 2 d = 22.6m d = 9.72m HW: pg 234 #1-5 Quantum