Physics - Kinematics

8. The average speed of Mrs. Wood's $32.0km$$32.0\ km$ trip to school was $20.0m/s$$20.0\ m/s$

   • A : How much time did it take Mrs. Wood to get to school ?

   $$\frac{20.0m}{1s}\ast \frac{0.001km}{1m}=\frac{0.02km}{1s}$$
   $$X\text{seconds}=\frac{32.0km}{\frac{0.02km}{1s}}$$
   $$X\text{seconds}=32.0km\ast \frac{1s}{0.02km}=1600.0\text{seconds}$$

   • B : She drove the first $20.0km$$20.0\ km$ at $25.0m/s$$25.0\ m/s$ , how much time did this take ?

   $$\frac{25.0m}{1s}\ast \frac{0.001km}{1m}=\frac{0.025km}{1s}$$
   $$X\text{seconds}=\frac{20.0km}{\frac{0.025km}{1s}}=800.0seconds$$

   • C : What speed did she drive in the last $12.0km$$12.0\ km$ ?

   $$\text{Remaining Time}=\text{Total Time}-\text{Time it took to travel the first part}=1600.0s-800.0s=800.0s$$
   $$\text{Speed}=\frac{12.0km}{800.0s}=\frac{0.015km}{1second}$$

9. Sean and Spencer start from $84.00m$$84.00\ m$ apart and start running towards each other. If Sean has a speed of $4.0m/s$$4.0\ m/s$ and Spenser has a speed of $3.0m/s$$3.0\ m/s$

   • A : How many seconds pass before they collide ?

     • You sort of have to just know to take their combined speed ,

       • When two objects are moving towards each other, their relative speed is the rate at which the distance between them decreases.

       • Imagine if one object were stationary and the other were moving towards it ,

         • The time to cover the distance would be the distance divided by the speed of the moving object.

       • However, if both objects are moving towards each other ,

         • the rate at which they "close the gap" is faster

         • this combined speed is simply the sum of their individual speeds.

   $$\text{Combined Speed}=\text{Sean’s Speed}+\text{Spenser’s Speed}=\frac{4.0m}{1s}+\frac{3.0m}{1s}=\frac{7.0m}{1s}$$
   $$\text{Collision Time}=\frac{84.0m}{\frac{7.0m}{1s}}=12.0seconds$$

   • B : What is the position at which they collide ?

   $$\text{Distance Sean Travels BEFORE Collision}=\frac{4.0m}{1s}\ast 12.0s=48.0m$$
   $$\text{Distance Spensor Travels BEFORE Collision}=\frac{2.0m}{1s}\ast 12.0s=36.0m$$

   ---

   • Now we have to find the position of collision relative to the starting point

   • We only really need to calculate the collision distance for one of the people , but we will do both just for confirmation

   • Lets say Sean starts at position/marker zero on our graph paper

     • and we say Sean moves towards +84

       • so we need a plus sign

     $$\text{Sean’s Position of Collision}=\text{Starting Position + Distance Sean Traveled BEFORE Collision}$$
     $$\text{Sean’s Position of Collision}=0.0m+48.0meters=48.0meters$$

   • And lets say Spencer starts at position/marker +84 on our graph paper.

     • and we say Spencer moves towards 0

       • so we need a minus sign

   $$\text{Spencer’s Position of Collision}=\text{Spencer’s Starting Position - Distance Spencer Traveled BEFORE Collision}$$
   $$\text{Spencer’s Position of Collision}=84.0m-36.0meters=48.0meters$$

Collision ( Example )

• Alex and Gabe start from $20.0m$$20.0\ m$ apart and start running towards each other. If Alex has a speed of $3.0m/s$$3.0\ m/s$ and Gabe has a speed of $2.0m/s$$2.0\ m/s$ :

  • A : How many seconds pass before they collide ?

    • we have again 2 objects approaching each other , so we need to find their combined speed

    $$\text{Combined Speed}=\frac{3.0m}{1s}+\frac{2.0m}{1s}=\frac{5.0m}{1s}$$
    $$\text{Collision Time}=\frac{20.0m}{\frac{5.0m}{1s}}=4.0seconds$$

  • B : What is the position at which they collide ?

  $$\text{Distance Alex Travels BEFORE Collision}=\frac{3.0m}{1s}\ast 4.0s=12.0m$$
  $$\text{Distance Gabe Travels BEFORE Collision}=\frac{2.0m}{1s}\ast 4.0s=8.0m$$

  • We could pick either person , but we only technically need 1 , as they both HAVE to be identical.

    • lets just say Alex starts at position zero , and moves to the right ,

      • making Gabe start at position 20 and moving to the left

  $$\text{Alex’s Position of Collision}=0.0m+12.0meters=12.0meters$$

Bikes ( Example )

• When the $10km/h$$10\ km/h$ bikes are $20km$$20\ km$ apart , a bee begins flying from one wheel to the other at a steady speed of $30.0km/h$$30.0\ km/h$ . When it gets to the wheel , it abruptly turns around and flies back to touch the first wheel , then turns around and keeps repeating the back-and-forth trip until the bikes meet , and "SQUISH !"

• How many kilometers did the bee travel in its total back-and-forth trips ?

  1. Find the time it takes the bikes to collide

     • we again have 2 objects moving towards each other , so we need the combined speed

     $$\text{Combined Speed}=\frac{10km}{1hour}+\frac{10.0km}{1hour}=\frac{20.0km}{1hour}$$
     $$\text{Collision Time}=\frac{\text{Distance}}{\text{Combined Speed}}$$
     $$\text{Collision Time}=\frac{20km}{\frac{20.0km}{1hour}}=1hour$$

  2. Calculate the distance the bee travels