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# Physics On the Road - Lesson 7

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### Physics On the Road - Lesson 7

1. 1. Speeding up, slowing down?
2. 2. Speeding up, slowing down Physics on the Road Lesson 7
3. 3. Constant acceleration When something gains speed at a constant rate this is called uniform acceleration. Lets begin with motion in a straight line vv∆vv+∆v∆v ∆t∆t∆t t=0 v v +∆v v +2∆v
4. 4. Traffic Incident A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible. Distance =150 m Initial velocity = 0 m/s Final velocity = 90 km/h = 25 m/s Time = 12 s Check 1: many cars advertise a 0-100 km/h of less than 10s Check 2: The car would need to gain 25/12=2.08m/s each of the 12 seconds or 2.08 m/s/s metres per second per second is the unit of acceleration and is abbreviated to ms-2.
5. 5. Traffic Incident cont… A policeman using a video camera and radar claims that a car starting from some lights reached 90 km/h when it passed a shop 150m away 12 s later. The driver claims this is impossible. Distance =150 m Initial velocity = 0 m/s Final velocity = 25 m/s Time = 12 s Car accelerating to 25m/s in 12s 0 5 10 15 20 25 0 1 2 3 4 5 6 7 8 9 10 11 12 Time /s speed/m/s ∆v ∆t acceleration = change of velocity time acceleration = 25 = 2.08 ms-2 12 avg. speed = change of distance time avg. speed = 150 = 12.5 ms-1 12 avg. speed = 0 + 25 = 12.5 ms-1 2 distance = avg. speed x time = 12.5 x 12= 150m
6. 6. Linking to distance time graphs  Draw this velocity time graph  Use your graph and the relationship between average speed and distance travel to complete a distance time table.  Plot the distance time graph Car accelerating to 25m/s in 12s 0 5 10 15 20 25 0 1 2 3 4 5 6 7 8 9 10 11 12 Time /s speed/m/s distance = avg. speed x time = 12.5 x 12= 150m Time /s Distance /m 12 150 Distance travelled is the area under a velocity time graph
7. 7. Kinematic equations acceleration = change of velocity time acceleration = final velocity – initial velocity time a = v – u t at= v – u v= u + at Car accelerating to 25m/s in 12s 0 5 10 15 20 25 0 1 2 3 4 5 6 7 8 9 10 11 12 Time /s speed/m/s Distance travelled is the area under the graph Area of triangle = ½ x base x height = ½ x time x change in velocity = ½ x 12 x 25 = 150 m First equation What if the object wasn’t originally stationary?
8. 8. More kinematic equations velocity time graph 0 1 2 3 4 5 6 7 8 0 1 2 3 4 time (s) velocity(m/s) The distance travelled is the area under the graph Area A + Area B A B u v u x t + ½ x t x (v-u) but… at = v – u so… s = ut + ½ at2 2nd equation
9. 9. Third equation of motion v= u + at t= v – u a average velocity= u + v 2 Distance = average velocity x time So…. s=u+v t 2 s = (u+v )x (v-u) 2 a s = v2 – u2 2a (u+v)(v-u)= v2 – u2 v2 = u2 + 2as3rd equation 1st equation
10. 10. Newton’s equations of motion v= u + at v2 = u2 + 2ass = ut + ½ at2 Use to compute final velocity from acceleration and time. Use to compute distance from acceleration and time. Use to compute Final velocity from acceleration and distance. v – final velocity (ms-1) t – time (s) u – initial velocity (ms-1) s –distance/displacement (m) a – acceleration (ms-2)
11. 11. Questions v= u + at v2 = u2 + 2as s = ut + ½ at2 A car travelling at 10 ms-1 accelerates at 3ms-2 for 5 seconds what is the car’s final velocity?. How far will the car travel whilst accelerating? A coin is dropped from the Centre Point tower in Sydney. The tower is 309m high. Acceleration due to gravity is 10 ms-2. What speed does the coin hit the ground? What assumption have you made? Q 1 a Q 1 b Q 2