Focus on 1-dimensional kinematics.
Initial Position:
At time ( t_1 = 0 ), position ( x_1 = -10.0 \text{ m} )
At time ( t_2 = 25.0 \text{ s} ), position ( x_2 = 40.0 \text{ m} )
Calculations:
Displacement:
( \Delta x = x_2 - x_1 = 40.0 \text{ m} - (-10.0 \text{ m}) = 50.0 \text{ m} )
Time Elapsed:
( \Delta t = t_2 - t_1 = 25.0 \text{s} - 0 = 25.0 \text{s} )
Distance (length of pool):
( d = 50.0 \text{ m} )
Average Velocity:
( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{50.0 \text{ m}}{25.0 \text{ s}} = 2.00 \text{ m/s} )
Average Speed:
( v_{avg (speed)} = \frac{d}{\Delta t} = \frac{50.0 \text{ m}}{25.0 \text{ s}} = 2.00 \text{ m/s} )
Initial Position:
At time ( t_1 = 0 ), position ( x_1 = 40.0 \text{ m} )
At time ( t_2 = 100.0 \text{ s} ), position ( x_2 = -10.0 \text{ m} )
Calculations:
Displacement:
( \Delta x = x_2 - x_1 = -10.0 \text{ m} - 40.0 \text{ m} = -50.0 \text{ m} )
Time Elapsed:
( \Delta t = t_2 - t_1 = 100.0 \text{s} - 0 = 100.0 \text{s} )
Distance:
( d = 50.0 \text{ m} )
Average Velocity:
( v_{avg} = \frac{\Delta x}{\Delta t} = \frac{-50.0 \text{ m}}{100.0 \text{ s}} = -0.500 \text{ m/s} )
Average Speed:
( v_{avg (speed)} = \frac{d}{\Delta t} = \frac{50.0 \text{ m}}{100.0 \text{ s}} = 0.500 \text{ m/s} ) (always positive)
Swimmer A:
Position Function: ( x_A = -10.0 + 2.0t )
Swimmer B:
Position Function: ( x_B = 40.0 - 0.50t )
Graphical Method:
Plotting ( x_A ) and ( x_B ) vs. time to find the intersection (meeting point).
Approximate meeting time at ( t = 20 \text{s} ) and position ( x = 30 \text{m} ) found through graphical analysis.
Setting Positions Equal:
Solve for ( t ) when ( x_A = x_B ):
( -10.0 + 2.0t = 40.0 - 0.50t )
Rearranging gives:
( 2.0t + 0.50t = 40.0 + 10.0 )
Final Calculation:
( 2.5t = 50.0 \Rightarrow t = 20.0 \text{ s} )
Validate positions:
( x_A = -10.0 + 2.0 \times 20.0 = 30.0 \text{ m} )
( x_B = 40.0 - 0.50 \times 20.0 = 30.0 \text{ m} )
Scenario:
Ball thrown upward with speed ( 15 \text{ m/s} )
Cliff height: ( 50 \text{ m} ) below.
Projectile Motion Specifications:
Initial Position ( x_0 = 50 , m )
Initial velocity ( v_0 = 15 , m/s )
Acceleration due to gravity ( a = -9.8 , m/s^2 )
Clicker Question:
Setup to solve for time ( t ) until reaching the base: using kinematic equations.
Scenario Description:
John driving at ( 126 \text{ km/h} ) encounters a truck going ( 90 \text{ km/h} )
Both vehicles decelerate at ( 6.00 \text{ m/s}^2 )
Initial distance apart: ( 144 , m )
Tasks:
A) Create velocity and position-time graphs to analyze collision likelihood.
B) If collision occurs, determine relative velocity at collision; if not, find stopping distance of both vehicles.
Setting Up:
Establish coordinate systems for both vehicles to simplify analysis.
Formula:
( v = v_0 + at )
Position Formula:
( x = x_0 + v_0t + \frac{1}{2}at^2 )
Graphs Visualization:
Sketching trajectories and velocities over time.
Vehicle Motion Contributions:
Compare changes in position over time for accuracy in collision estimation.
Visualize the motion with graphs to ascertain the impact of deceleration.
Graphs:
Important to illustrate the position changes of both vehicles over time to identify collision points visually.
Detailed Time Analysis:
Report the position of each vehicle at specific time intervals leading up to their stop.
Example calculations at times ( t = 15, 25, 35 ) seconds.
Collision Confirmation:
Based on time-dependent position analysis, confirm whether collision occurred.
Graphical vs. Analytical Solutions:
Emphasize the need for solving equations analytically for precision rather than relying solely on graphical estimates.
Conclusion:
Always verify findings through systematic computation alongside visual aids.