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Displacement - time

Displacement - time. An object travelling along a straight line has two possible motions - forward/backwards Sketch a displacement time graph for a journey where you return to your starting position. Challenge: sketch a displacement time graph for a swinging pendulum.

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Displacement - time

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  1. Displacement - time • An object travelling along a straight line has two possible motions - forward/backwards • Sketch a displacement time graph for a journey where you return to your starting position. • Challenge: sketch a displacement time graph for a swinging pendulum. • Remember time is on the x-axis

  2. A LEVEL PHYSICSYear 1 • I can use and rearrange the suvat eqautions to identify an unknown variable (Grade C). • I can use speed – time graphs to determine the distance travelled (Grade B). • Be able to use the SUVAT equations to solve complex linear problems (Grade ). A* Acceleration and SUVAT A B C

  3. To start Using the equation below, state and prove the units for a. v = u + at Final velocity Initial velocity

  4. The knowledge

  5. The knowledge a = (v-u)/t v – u = at Acceleration is the rate of change of velocity.

  6. The knowledge vav = s/t s = vav x t vav = (u + v)/2 s = (u + v)/2 x t If acceleration is constant, the average velocity is just the average of the initial and final velocities.

  7. The knowledge s = (u + v)/2 x t s = (u + (u + at))/2 x t s = (2u + at)/2 x t s = (2ut + at2)/2 s = ut + 1/2 at2 By combining equations (1) and (2) you can eliminate v. Substitute v = u + at

  8. The knowledge a x s as = (v-u)/t x (u + v)/2 x t as = (v – u)(u + v)t/2t as = (vu + v2 - u2 –vu)/2 as = (v2 - u2)/2 2as = (v2 - u2) v2 = u2 + 2as By combining equations (1) and (2) you can eliminate t. Multiply a = (v-u)/t and s = (u + v)/2 x t

  9. Using velocity time graphs to find displacement Displacement is the area under the line between the start and time t. Constant velocity Eq.2: s = (u + v)/2 x t Again displacement on the graph is the area beneath the line. Trapezium of base t and average height corresponding to average speed (u + v)/2. Constant acceleration Changing acceleration

  10. Using velocity time graphs to find displacement Let v represent the velocity at time t and v+dv represent the velocity a short time later at time t+dt. The velocity dv is small compared with the velocity v, the displacement ds in the short time interval dt is vdt. On the graph this is the shaded strip area – base of dt and height v (ds = vdt). Now consider the whole area under the line in strips of similar width – the total displacement from the start to t. This is the sum of the area of every strip. Changing acceleration

  11. Have a go • A parachutist jumps out of an aeroplane and accelerates uniformly at 9.8 ms-2 for ten seconds: • How far does she travel? • What is her final velocity? • A car accelerates uniformly from 5 ms-1 to 20 ms-1 over a distance of 500m • How long does this take? • What is the car’s acceleration?

  12. Answers • 1. s = 490m2. v = 98m/s • 1. t = 40s2. v = 0.375m/s

  13. Quick Quiz

  14. B/C Task A/A* Task • Complete and mark Practice Questions – Application Pg.124 of your textbook. • Complete Summary Questions Pg.124 of AQA Physics. • Complete and mark Practice Questions – Application Pg.124 of your textbook. • Complete Summary Questions Pg.124 of AQA Physics. • Pg. 136 of AQA Physics.

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