Mechanics 1

• Mechanics involves the study of motion and its causes.

Relevance Demonstrated Through Exam Question

• A question from a 2005 physics final exam highlights the importance of understanding mechanics.
• The problem involves two cars, one moving at a constant velocity and the other accelerating.
• The question asks for the speed of the accelerating car when it passes the car moving at a constant speed.
• Less than 50% of students answered correctly.

Kinematics Learning Goals

• Kinematics is the description of motion, and why it's important in physics.
• Fundamental kinematic quantities are time, distance, speed, and acceleration.
• Vector quantities include displacement, velocity, and acceleration.
• Kinematics also covers average and instantaneous quantities.
• Relative displacement, velocity, and acceleration are part of kinematics.

Understanding Motion

• Quantities are needed to describe motion adequately.
• Motion involves a change in position with time.
• Time and position must be measured to analyze motion.
• Rates of change, including position and the rate of change of the rate of change, are important.

Time, Distance, and Speed Defined

• Calculating distance traveled over time by an object.
• A car moving at 5 m/s travels 15 meters in 3 seconds.
• Speed is the rate of change of an object's position over time, expressed as v = Δx/Δt, where Δ means "change in."
• To find the distance traveled at a constant speed, use Δx = vΔt.
• For a car traveling at 5 m/s for 3 seconds, the distance covered (Δx) is 15 meters.

Average Speed Explained

• Average speed is total distance traveled divided by total time.
• Average speed is expressed as V(av) = Δx/Δt.
• Average speed is not the speed at a particular time.
• At constant acceleration, average speed = (V(f) + V(i)) / 2.

Instantaneous Speed Defined

• Instantaneous speed is the speed of an object at a particular moment.
• As Δx and Δt decrease, average speed approaches instantaneous speed.

Example of Average Speed

• Driving 300 km to Christchurch in 5 hours results in an average speed of 60 km/h.
• The instantaneous speed varies during the trip.

Understanding Acceleration

• Acceleration is the rate of change of speed.
• Acceleration is expressed as a = Δv/Δt.
• The change in speed for a given uniform acceleration is found using Δv = aΔt.
• A car accelerating from rest at 2 m/s² for 4 seconds will have a final speed of 8 m/s.

Train Speed Example

• A train's speed increases uniformly from 3 m/s to 11 m/s over 100 seconds, resulting in an average speed of 7 m/s.
• The equation provides the average speed, V(av) = (V(f) + V(i)) / 2.
• The train travels 700 meters in 100 seconds (Δx = V(av)Δt).

Uniform Acceleration Detailed

• For an object with uniform acceleration, Δx = V(av)Δt.
• Δx = 1/2 * (Vi + Vf) * Δt which can be expressed as Δx = ViΔt + 1/2 aΔt².
• If the object starts from rest, Δx = 1/2 aΔt².

Vector Quantities Specified

• Displacement: Vector describing distance and direction to an object, exemplified by "3 m north" or "4 m east."
• Velocity: Rate of change of displacement, illustrated by "12 km/h south" or "10 m/s NE."
• Acceleration: Rate of change of velocity.
• Deceleration: Acceleration opposing the velocity vector.

Adding Velocities: Understanding Relative Motion

• When an airplane flies at 100 km/h in still air with a 30 km/h wind, its speed relative to the ground changes.
• If the wind is in the same direction, the resultant speed is 130 km/h.
• If the wind is in the opposite direction, the resultant speed is 70 km/h.
• A bird flying north at 3 m/s with a 4 m/s easterly wind has a resultant velocity of 5 m/s at 37° N of E.

Acceleration Due to Gravity

• Galileo discovered objects falling freely towards Earth accelerate equally, a = g = 9.8 m/s² ≈ 10 m/s².
• 'g' signifies acceleration due to gravity.
• Velocity changes by Δv = gΔt under gravity.
• After 5 seconds of free fall from rest, Δv = 10 m/s² * 5 s = 50 m/s.

Free Fall Calculations

• A cricket ball dropped from a plane falls 125 m in 5 seconds.
• Uses formulas for constant acceleration to determine free fall distance.

Projectile Motion Explained

• A cricket ball thrown upwards at 30 m/s will reach a height of 45 meters.
• Formulas used are Vf = Vi + aΔt, Δx = V(av)t.
• Takes 3 seconds to reach max height.

Example of Time to Reach the Ground

• The ball takes same time (3 s) to come down as to go up.
• Starts with Vi=0, Δx=45 m, g=10 m/s using the formula Δx = (1/2)gΔt².
• Downward velocity at ground is the same as the launch (30m/s).

Vertical and Horizontal Motion

• Vertical and horizontal motions are considered independent.
• Objects fall at the same rate.

Ball Flight Trajectory

• If horizontal velocity = 10 ms-1, and the flight is independent, then a parabolic motion occurs.

Relative Motion

• A polar bear is 200m behind a man running at 7 m/s.
• The polar bear is running at 11 m/s to the man's car.
• The bear's velocity relative to the man is 4 m/s.
• It will take the bear 50secs to catch the man.