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Topic 1.3 Extended B - Components of motion Up to now we have considered objects moving in one dimension. However, most objects move in more than one.

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Presentation on theme: "Topic 1.3 Extended B - Components of motion Up to now we have considered objects moving in one dimension. However, most objects move in more than one."— Presentation transcript:

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2 Topic 1.3 Extended B - Components of motion

3 Up to now we have considered objects moving in one dimension. However, most objects move in more than one dimension. For example, consider the ball shown here: Motion in Two Dimensions 3-1 Components of motion We can sketch in our x and y for successive snapshots to obtain an idea of the different velocities the ball has at different times: x is in YELLOW. y is in RED. We can also sketch in the displacement d of the ball at each time interval (in GREEN). Let's examine one time interval in detail: x y d y FYI: The displacement vector gives the direction of the motion

4 From the Pythagorean Theorem we can find the value of d if we know x and y: d 2 = x 2 + y 2 Topic 1.3 Extended B - Components of motion x y d y d = x 2 + y 2 Magnitude of a 2D displacement If we know the time interval t between snapshots, we can find the velocity of the ball simply by dividing the displacements shown above by t. The proportions of our triangle will not change. vxvx vyvy v vyvy Thus v = v x 2 + v y 2 Magnitude of a 2D velocity Each triangle gets a good name: displacement triangle velocity triangle

5 We call the v x the horizontal component of the velocity. Topic 1.3 Extended B - Components of motion vxvx vyvy v vyvy horizontal component We call the v y the vertical component of the velocity. vertical component From trigonometry we know there is a relationship between the sides of a triangle, and the angle : opp hyp adj hyp opp adj hypotenuse adjacent opposite θ trigonometric ratios s-o-h-c-a-h-t-o-a v v x = v cos θ v y = v sin θ v vxvx vyvy v sin θ = cos θ = tan θ = vxvx vyvy

6 Suppose we know the velocity of the ball is 25.0 m/s at an angle of 30° with respect to (wrt) the positive x-axis. Topic 1.3 Extended B - Components of motion vxvx vyvy v vyvy What is v x the horizontal component of the velocity? v x = v cos θ v y = v sin θ v x = v cos θ v x = (25.0 m/s)cos 30° v x = 21.7 m/s What is v y the vertical component of the velocity? v y = v sin θ v y = (25.0 m/s)sin 30° v y = 12.5 m/s FYI: You can check your results by squaring each answer, summing, and taking the square root. What should you get?

7 Sometimes we know the components of the velocity, and want to find the magnitude and the direction: Topic 1.3 Extended B - Components of motion vxvx vyvy v vyvy Suppose v x = 30.0 m/s. Suppose v y = 40.0 m/s. Then v = v x 2 + v y 2 v = 30 2 + 40 2 v = 50.0 m/s magnitude of v opp adj tan θ = vxvx vyvy = 40 m/s = 30 m/s and so that θ = tan -1 4343 = 53.1° direction of v

8 Sometimes we know the formulas for the components of the velocity of a ball, and want to find the magnitude and the direction of the velocity at a particular time: Topic 1.3 Extended B - Components of motion Suppose v x = 30.0 (measured in m/s). Suppose v y = 40.0 - 5t (v y in m/s, t in s) Then what is the velocity at t = 2 s? v = v x 2 + v y 2 v = 30 2 + 30 2 v = 42.4 m/s magnitude of v opp adj tan θ = vxvx vyvy = 30 m/s = What is the direction of the ball at this instant? so that θ = tan -1 (1) = 45.0° direction of v v x = 30.0 m/s v y = 40 - 5(2) v y = 30.0 m/s


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