Sierzega: Kinematics 10 Page 1 of 14 10.1 Hypothesize (Derive a Mathematical Model) Graphically we know that the area beneath a velocity vs. time graph line represents the displacement of an object. For constant velocity objects we developed the following model: x(t) = x o + v o t When the velocity is changing as it did in the experiment with the falling ball we can find the area under the velocity vs. time graph line with a little geometry. The area of a triangle is: ½ (height)*(base) The base of our triangle is the time interval Δt (which is the same as t, while our height is the change in velocity Δv. ½ ΔvΔt If substitute Δv for an expression with the acceleration a a = Δv/Δt aδt = Δv We get the following expression for the area under the velocity vs. time graph line, which corresponds to the displacement of an accelerating object: Which simplifies to: ½ aδt 2 ½at 2
Sierzega: Kinematics 10 Page 2 of 14 If we take Δt = t - t i and set t i to 0. If our object has an initial velocity then the y-intercept of the velocity vs. time graph shifts. For example: ½at 2 v o v o t t The displacement then becomes the combined area of the top triangle and the rectangle beneath plus any initial displacement. The generalized equation for displacement then is: This assumes constant acceleration. x(t) = x 0 + v 0 t + ½at 2
Sierzega: Kinematics 10 Page 3 of 14 10.2 Test Your Ideas with Phet Simulations Scenario 1: a) x(t) = (-8 m) + ½(0.75 m/s 2 )t 2 b) c) The functions, written descriptions and graphs are all consistent with each other. The object starts at rest and accelerates at a constant rate resulting in a velocity vs. time graph that should start at zero and have a steady slope. The position vs. time graph on the other hand should start at -8 m (the tree) and increase exponentially, meaning the slope of the line should get steeper and steeper with every second as the velocity gets larger and larger. d) x(t) = (-8 m) + ½(0.75 m/s 2 )t 2 x(t) = (-8 m) + ½(0.75 m/s 2 )(5.0s) 2 x = 1.4 m e) The prediction and outcome agree.
Sierzega: Kinematics 10 Page 4 of 14 Scenario 2: a) x(t) = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 b) c) Everything is consistent. The position vs. time graph intercepts the y-axis at -7 m and increases exponentially, as expected with accelerated motion. The velocity vs. time graph intercepts the y-axis at 0.75 m/s (his initial velocity) and increases at a constant rate of 0.2m/s 2. d) x(t) = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 0 m = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 Quadratic Equation: c + bx + ax 2 c = -7 m
Sierzega: Kinematics 10 Page 5 of 14 b = 0.75 m/s a = 0.2 m/s 2 x = t t = 5.41 s, -12.91 s Because time cannot be negative: t = 5.41 seconds e) The prediction and outcome agree.
Sierzega: Kinematics 10 Page 6 of 14 Scenario 3: a) x(t) = (5 m) + (-1.0 m/s)t + ½(-0.5 m/s 2 )t 2 b) Everything is consistent. The position vs. time graph intercepts the y-axis at 5 m and goes towards zero with the slope getting more and more negative, as expected with accelerated motion where the object is speeding up in the negative direction. The velocity vs. time graph intercepts the y-axis at -1.0 m/s (his initial velocity) and decreases at a constant rate of -0.5m/s 2. c) Find the time that the man is moving at 5 m/s: v(t) = v o + at t = (v - v o )/a t = [(-5 m/s) (-1 m/s)]/(-0.5 m/s 2 ) t = 8 s
Sierzega: Kinematics 10 Page 7 of 14 Find the position at 8 s: x(t) = (5 m) + (-1.0 m/s)t + ½(-0.5 m/s 2 )t 2 x(t) = (5 m) + (-1.0 m/s)(8 s) + ½(-0.5 m/s 2 )(8 s) 2 x = -19 m d) The prediction and outcome agree.
Sierzega: Kinematics 10 Page 8 of 14 Scenario 4 a) x(t) = (8 m) + (-7.0 m/s)t + (1.0 m/s 2 )t 2 b) c) Everything is consistent. The position vs. time graph intercepts the y-axis at 8 m and goes towards zero with the slope getting less and less steep, as expected with accelerated motion where the object is slowing down in the positive direction. The velocity vs. time graph intercepts the y-axis at -7.0 m/s (his initial velocity) and increases at a constant rate of -1.0 m/s 2, meaning his speed is decreasing at 1.0 m/s 2. d) Find the time that the man is moving at 0 m/s: v(t) = v o + at t = (v - v o )/a t = [(0 m/s) (-7 m/s)]/(1.0 m/s 2 ) t = 7 s
Sierzega: Kinematics 10 Page 9 of 14 Find the position at 7 s: x(t) = (8 m) + (-7.0 m/s)t + (1.0 m/s 2 )t 2 x(t) = (8 m) + (-7.0 m/s)(7.0 s) + (1.0 m/s 2 )(7.0 s) 2 x = -16.5 m e) The prediction and outcome agree.
Sierzega: Kinematics 10 Page 10 of 14 10.3 Compare and Contrast a) A car is pulling out of the driveway moving in the negative direction at a constant velocity. v 4 v 3 v 2 v 1 Δv 12 Δv 23 Δv 34 +x b) A car is passing another car on the highway, speeding up to get past. v 1 v 2 v 3 v 4 Δv 12 Δv 23 Δv 34 +x c) A car is slowing down at a red light. v 1 v 2 v 3 v 4 Δv 34 Δv 12 Δv 23 +x d) A car is passing another car on the highway, speeding up to get past. The coordinate system has been changed. v 4 v 3 v 2 v 1 Δv 34 Δv 23 Δv 12 +x
Sierzega: Kinematics 10 Page 11 of 14 10.4 Represent and Reason Answers will vary depending on coordinate system. a) Sketch 12 m/s -6 m/s 2 0 meters 20 meters +x b) Motion Diagram v 1 v 2 v 3 v 4 c) Position vs. Time Δv 12 Δv 23 Δv 34 +x
Sierzega: Kinematics 10 Page 12 of 14 d) Velocity vs. Time e) x(t) = (12 m/s)t + ½(-6 m/s 2 )t 2 v(t) = (12 m/s) + (-6 m/s 2 )t f) Time when car stops: 0 m/s = (12 m/s) + (-6 m/s 2 )t t = (-12 m/s)/(-6 m/s 2 ) = 2 s Position when car stops: x = (12 m/s)(2 s) + ½(-6 m/s 2 )(2 s) 2 = 12 m The mystery machine stops in 12 meters, or 8 meters from the stop sign. It takes 2 seconds to come to a complete stop.
Sierzega: Kinematics 10 Page 13 of 14 10.5 Reason and Represent Answers will vary based on the reference frame chosen a) 6.0 m/s 8.0 s later 9.75 m/s 0 meters +x b) v 1 v 2 v 3 v 4 v 5 v 6 v 7 Δv 12 Δv 23 Δv 34 Velocity vs. Time Δv 45 Δv 56 Δv 67 +x d) Distance Traveled before Acceleration x = (6.0 m/s)(8.0 s) = 48 m Distance Traveled after Acceleration: ( vf vi) a t a = (9.75 m/s 6.0 m/s)/(5.0 s) = 0.75 m/s 2 x(t) = (6.0 m/s)t + ½(0.75 m/s 2 )t 2 x = (6.0 m/s)(5.0 s) + ½(0.75 m/s 2 )(5.0 s) 2 = 39 m Total Distance: 48 m + 39 m = 87 m
Sierzega: Kinematics 10 Page 14 of 14 10.6 Regular Problem a) A car starts at a velocity of 4m/s in the positive direction and accelerates uniformly for 6 seconds until it is at a speed of 22m/s. b) Acceleration of the car: ( vf vi) a t a = (22 m/s 4 m/s)/(6 s) a = 3 m/s 2 Displacement of the car is equal to the area under the graph line: x(t) = (4 m/s)t + ½(3 m/s 2 )t 2 x = (4 m/s)(6 s) + ½(3 m/s 2 )(6 s) 2 = 78 m c) If the object traveled at a constant velocity of 4 m/s (the velocity at t = 0 s) for 6 s the total displacement would be 24 m. This makes sense since it is less than the displacement for the accelerating object. 24 m is equal to the first term of the x(t) function. It is the displacement if a = 0 m/s 2