v a =, to calculate acceleration. t You had also re-arranged it as v = a t using basic Algebra (or cross multiplying) Velocity

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1 1 GRADE 1 SCIENCE ACCELERATION DISTANCE VS VELOCITY **MORE IMPORTANTLY: VELOCITY VS BRAKING DISTANCE** Name: Date: 1. You have studied the formula v a =, to calculate acceleration. t You had also re-arranged it as v = a t using basic Algebra (or cross multiplying) 2. So you see how the change in velocity depends on the time. The longer time you push something the faster it goes! WOW! 3. If you push something at a constant acceleration for twice the time, velocity gets twice as fast! It is a linear relationship Velocity vs Time a = 2 Velocity Time But how does velocity depend on the distance you accelerate it? It turns out the secret formula is that d depends on the velocity squared! This assignment is another way to better explore pages 27 to 3 in your In Motion notes. To keep it simple, the secret formula is v 2 = 2ad. If you accelerate something (from rest) over a certain distance, the square of the velocity goes up with the distance you accelerate the body. 5. Confused? Lets do a table! A table is one wonderful way to better see a pattern! 6. Calculate the table for a uniform [constant] acceleration of a = 1 m/sec/sec or 1m/sec 2. You complete the last three columns. d [m] v 2 =2ad v [m/sec] Gr11Ess_DistanceVelocity.doc Revised:

2 2 7. Can you see the pattern? You may need to recall the square root,,from Grade 8 or 9 Math to un-square a value. 8. What happens to the velocity when you quadruple (x4) the distance you accelerate your object through? Of course you follow the same pattern when decelerating! (ie: trying to stop a car!). So going twice the speed takes four times the distance to stop! 9. Try graphing the table above to get a visual of the pattern, you plot the last three points from the table: Velocity vs Distance Accelerated at 1 m/sec Distance Accelerated [m] So that is an interesting pattern but not too many drivers ask themselves how far do I need to accelerate to get to a certain speed? 11. Often drivers are more interested in the backwards question, the inverse question, what distance will it take me to stop if I decelerate? That is: how far will I skid when I put on the brakes! Just think backwards; or sideways, on the graph. Flip it sideways and around.

3 3 7 Velocity vs Distance Accelerated at 1 m/sec Distance Accelerated [m] Distance vs Velocity Accelerating at 1 m/sec 2 2 Distance [m] Think about what this means!!!! Now that should scare you! Going twice the speed takes four times the braking distance to stop! You may want to say that to yourself 1 times before you start out on a slippery winter drive! So going half the speed would take a quarter the braking distance. 13. And that does not allow for your reaction time! That distance would be added on too! There is a notional formula for your braking distance depending on how slippery the road is or how good your brakes are. 14. The braking distance of car, once the brakes are applied, is a function of slipperiness of the road and speed. In fact, the stopping actually depends on the square of the speed! (Really! Double your speed: quadruple your stopping distance! Halve your speed, gets a quarter of your braking distance. Or quadruple the damage to your head when you hit a telephone pole). The formula for the braking distance then is like this: D =.1*K*v In this formula, D is distance to stop in meters. The initial speed when you start braking, v, is in km/hour. The K is an extra coefficient to allow for slipperiness and friction with the road. When K is 1 the road is normal and dry. When K is more than 1 (K>1) the road is slippery. On ice, K, can be as high as What braking distance does it take for the car to stop if initial speed, v, is 5 km/hr and the roads are dry, a. K=1.? (hint: evaluate, plug in). b. How about if wet and slippery K = 2?

4 4 17. What braking distance does it take for the car to stop if initial speed, v, is 1 km/hr and the roads are dry, a. K = 1.?. b. How about if wet and slippery K = 2? 18. Complete the blanks in the stopping distance table: Table of Stopping Distance, D, for various Friction coefficients, K, and various velocities, V. V K D =.1*K*v 2 Value of D 3 1 D = 6 1 D = 9 1 D = 3 2 D = 6 2 D = 9 2 D = 3 4 D = 6 4 D = 9 4 D = Now how would you do it backwards; solve for a speed that gives you a certain skid distance??? 4 D = 8 meters 8 meters?? 1.75 D = 4 meters 4 meters

5 5 19. Now graph below the data from the table we calculated above!. (You will need to graph three separate parabolic curves, one for each given K ) Skid Distance for Various Coefficients of Friction skid distance, D [m] speed, v [km/h] Your graph should resemble this: Notice how by doubling your speed you quadruple your stopping distance! Graph all three curves on the calculator! Check the table, check the graph. If you have a graphing tool! Go for it! skid distance, D [m] Skid Distance for Various Coefficients of Friction speed, v [km/h] Advanced Question Solving a Quadratic. How fast was someone going if their skid mark was 8 meters long on a dry day? This requires a bit of simple algebra!

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