SPH3U/4C Unit 1 MotionLESSON #6: GRAPHING NON - UNIFORM MOTION

Date:

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Accelerated MotionAcceleration is defined as the rate of change of velocity.

If a moving object accelerates, its motion is non-uniform.

The acceleration is determined using the formula:

�⃑� =�⃑� − �⃑�∆𝑡

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Accelerated Motion ExampleEg.#1. A ball thrown into the air has an initial velocity of 50. m./s.[up]. If, after 10. seconds, its velocity is 50. m./s.[down], what is its acceleration?

�⃑� =�⃑� − �⃑�∆𝑡

�⃑� =50.𝑚. 𝑠.⁄ 𝑑𝑜𝑤𝑛 − 50. 𝑚. 𝑠.⁄ [𝑢𝑝]

10. 𝑠.

�⃑� =50.𝑚. 𝑠.⁄ 𝑑𝑜𝑤𝑛 + 50.𝑚. 𝑠.⁄ [𝑑𝑜𝑤𝑛]

10. 𝑠.

�⃑� = 10.𝑚.𝑠. [𝑑𝑜𝑤𝑛]

Therefore the ball had an acceleration of 10. m./s.2[down]

�⃑� = 50.𝑚. 𝑠.⁄ [𝑢𝑝]

�⃑� = 50.𝑚. 𝑠.⁄ [𝑑𝑜𝑤𝑛]

∆𝑡 = 10. 𝑠.

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�⃑� =?

Accelerated Motion ExampleEg.#2. A go-cart is traveling up a hill at 5.0 m/s when, suddenly, the engine gives out. The cart now experiences an acceleration of 2.0 m/s2.[down the hill]. What is its velocity after 3.0 seconds?

�⃑� =�⃑� − �⃑�∆𝑡

�⃑�∆𝑡 = �⃑� − �⃑�

�⃑� + �⃑�∆𝑡 = �⃑�

�⃑� = 5.0𝑚. 𝑠.⁄ 𝑢𝑝 − 2.0𝑚.𝑠. 𝑢𝑝 × 3.0𝑠.

�⃑� = −1.0𝑚. 𝑠.⁄ 𝑢𝑝

�⃑� = 1.0𝑚. 𝑠.⁄ 𝑑𝑜𝑤𝑛The final velocity of the go-cart is 1.0m./s.[down the hill]

�⃑� = 5.0𝑚. 𝑠.⁄ 𝑢𝑝 + 2.0𝑚.𝑠. 𝑑𝑜𝑤𝑛 × 3.0𝑠.

�⃑� = 5.0.𝑚. 𝑠.⁄ [𝑢𝑝]

�⃑� = 2.0𝑚.𝑠. [𝑑𝑜𝑤𝑛]

∆𝑡 = 3.0𝑠.

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�⃑� =?

�⃑� = 5.0𝑚. 𝑠.⁄ 𝑢𝑝 − 6.0𝑚. 𝑠.⁄ 𝑢𝑝

Accelerated Motion ExampleEg.#3. A model rocket traveling upward at 40. m./s. runs out of fuel. It then experiences an acceleration of 9.8 m./s.2 towards Earth. How long after this happens will it be traveling back towards Earth at 50. m./s.?

�⃑� = 40.𝑚. 𝑠.⁄ [𝑢𝑝]

�⃑� = 50.𝑚. 𝑠.⁄ [𝑑𝑜𝑤𝑛]�⃑� = 9.8𝑚.

𝑠. [𝑑𝑜𝑤𝑛]

∆𝑡 =?

�⃑� =�⃑� − �⃑�∆𝑡

�⃑�∆𝑡 = �⃑� − �⃑�

∆𝑡 =�⃑� − �⃑�

�⃑�

∆𝑡 =50.𝑚. 𝑠.⁄ 𝑑𝑜𝑤𝑛 − 40. 𝑚. 𝑠.⁄ [𝑢𝑝]

9.8𝑚.𝑠. [𝑑𝑜𝑤𝑛]

∆𝑡 =50.𝑚. 𝑠.⁄ 𝑑𝑜𝑤𝑛 + 40.𝑚. 𝑠.⁄ [𝑑𝑜𝑤𝑛]

9.8𝑚.𝑠. [𝑑𝑜𝑤𝑛]

∆𝑡 = 9.2s.

Therefore, after 9.2 seconds the rocket will be travelling back towards Earth at 50.m./s.

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Graphing Non-Uniform MotionIn  most  of  the  examples  we’ve  looked  at  so  far,  the  graph  of  the  position  versus  time  was  a  straight line.

If the motion of an object is non-uniform, its position vs. time graph will not be a straight line.

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Curved Line SlopesWhen working with a curved line graph, we can take the slope in two ways:◦ Secant line slope:

◦ Joins two points together ◦ Is used to find the average slope or rate of change between the two points on a graph

◦ Tangent line slope◦ Lines up with only one point on

the graph◦ Is used to find the instantaneous

slope at a single point on a graph

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Instantaneous SlopeTo find the instantaneous slope, we must draw a tangent line to the graph at the point we are interested in.

Eg.#4. Find the instantaneous slope of the graph below at t = 5.0 seconds.

𝑚 =𝑦 − 𝑦𝑥 − 𝑥

=8.5𝑚. 𝐸 − 4.0𝑚. [𝐸]

7.6𝑠.−3.0𝑠.

= 1.2 𝑚. 𝑠.⁄ [𝐸]

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The instantaneous velocity at t = 5.0 seconds is 1.3 m./s.[E]

Average SlopeEg.#5. Find the average velocity of the car with the position vs. time graph shown for the time between 2.0s. and 8.0s.

𝑚 =𝑦 − 𝑦𝑥 − 𝑥

=7.5𝑚. 𝐸 − 2.3𝑚. [𝐸]

8.0𝑠 − 2.0𝑠.

=5.2𝑚. 𝐸6.0𝑠.

= 0.87𝑚. 𝑠.⁄ [𝐸]

Therefore the average velocity of the car is 0.87m./s.[E]

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Graphing Non-Uniform MotionEg.#6. Given the graph of non-uniform motion below, describe the motion.

1. The squirrel begins with a large velocity (slope) which decreases uniformly (decreasing slope) until t=5.0s.

2. At t=5.0s, the squirrel stops for 2.5 seconds.

3. The squirrel moves West with a uniform velocity for 2.5 seconds.

4. The graph ends with the squirrel located at 20.0m[E.] of the starting position ①

②③

④

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