Kinematics Question Notes

Kinematics Questions 1. The velocity-time graph for an object moving north: * The object is increasing its velocity with a constant acceleration. 2. A boat leaves a marina and sails 2 km [E], then 2 km [E], then 2 km [N], and finally 2 km [W], taking a total time of 2 hr for the trip. What is the average velocity of the boat? Vavg = total Δd / total Δt = 0 km / 2 h = 0 km/h 3. A ship started from rest and accelerated at 0.20 m/s². How long would it take to reach 5 m/s? Given: Vi = 0 m/s Vf = 5 m/s a = 0.20 m/s² Δt = ? Vf = Vi + at 5 = 0 + 0.20t 5 = 0.20t 25 = t 4. An object held 8 m above the ground and released. (position vs time graph) d(m) 8 6 4 2 0 0 2 4 6 t(s) 5. Linda walks 300m [E], then 300m [N]. What direction must she walk to return directly to her starting point? [Diagram: A right-angled triangle with sides 300m each] Answer: [W 45° S] 6. An object undergoes a displacement (Δd₁) followed by a second displacement (Δd₂). When using the Scale Diagram Method to add the vectors, you must... Answer: add them tail to tail 7. To find the total displacement for an object moving in two dimensions can be found using: scale diagram method cosine and sine law method perpendicular components of a vector method 8. Average speed is always greater than or equal to the magnitude of the average velocity for an object moving in two dimensions.   9. Uniform and Non-Uniform Velocity Examples: Uniform Velocity A helicopter climbing upwards at a constant speed in a straight line A motorcycle driven up a hill in a straight line at a constant speed Non-Uniform Velocity A bicyclist accelerating down a hill in a straight line A race car driving around an oval track at a constant speed A skydiver free falling after jumping from an airplane 10. Acceleration due to gravity seems like a straight forward concept, but falling bodies fall so quickly that it is impossible for the human eye to really present a clear picture of the motion. How can the understanding of acceleration due to gravity be made? To understand acceleration due to gravity, I started by throwing an eraser into the air and noticed that it won't just float upwards, but instead falls downward due to gravity. I connected this real-life example to a sky diver. Since I know from doing dynamics questions that free fall is non-uniform, so the diver is accelerating towards the ground. I also connected this to my knowledge of Earth's gravitational field. I know that the closer we are to the Earth, the stronger we feel it's pull, bringing us downward towards Earth's core. So as the diver and the eraser move closer to the ground, it feels the earth's pull more strongly and therefore move towards the ground at a quicker pace. This is how I understood acceleration due to gravity. 11. A position-time graph for an object moving west, with a corresponding velocity-time graph (with calculations) [Image of position-time graph and velocity-time graph] Calculations: 1. v = Δd/Δt = (4-0)/(4-0) = 1 m/s 2. v = Δd/Δt = (8-8)/(8-6) = 0 m/s 3. v = Δd/Δt = (10-8)/(12-8) = 0.5 m/s 12. A man was walking at 1.5 m/s when he saw a rabbit and accelerated at 3.2 m/s^2 in a straight line for 3.2 s. What was his final speed? Given: Vi = 1.5 m/s a = 3.2 m/s^2 t = 3.2 s Vf = ? Vf = Vi + at Vf = 1.5 + (3.2)(3.2) Vf = 12.04 m/s 13. A light aircraft flew 2.5 x 10^2 km [W 30°S]. Determine the components of the displacement. [Image of a right-angled triangle with sides dx, dy, and hypotenuse 2.5 x 10^2 km] cosθ = dx/hypotenuse cos30° = dx / 2.5 x 10^2 dx = 2.165 x 10^2 km sinθ = dy/hypotenuse sin30° = dy / 2.5 x 10^2 dy = 1.25 x 10^2 km 14. A pilot diverted an airliner 20 km [E 45° N] to avoid a thunderstorm cell, and then flew 20 km [W 45° N] to get back on track. If the diversion lasted a total of 3 minutes, what is the average velocity of the airliner during the diversion? [Image of a diagram showing the displacement vectors] Average Displacement: d = √(20^2 + 20^2) = 28.28 km Time/Average Time: 3 minutes x (1 hour/60 minutes) = 0.05 hr Average Velocity: Vavg = Δd/Δt = 28.28 km / 0.05 hr = 565.6 km/h