Speed and Velocity Speed is a measure of how fast an object moves. Speed = distance scalar time interval Velocity is the rate at which an object changes position. velocity = change in position time interval vector
Constant velocity V An object with a constant velocity moves at a constant speed in a constant. direction Average Velocity V Average velocity is used if an object undergoes a change in speed and or direction during the time interval.
You can check out this animation of varying speeds: http://www.physicsclassroom.com/mmedia/kinema/trip.cfm URL Car must slow to a stop at red light. Total Car can continue driving (maybe speeding up from rest) at a green light. A car moving along this stretch of road would not be able to move at a constant velocity. Must stop at the stop sign and any red light. So, one could calculate the average velocity for the entire length of the street shown. d 1 = 0 miles [E] d 2 = 5.0 miles [E] Δt = 0.20 h step 1: list givens ah, an American site! Δd = d 2 - d 1 = 5.0 miles - 0 miles = 5.0 miles [E] step 2: equations step 3: substitutions v = Δd Δt = 5.0 miles 0.20 h step 4: round to correct number of significant digits = 25 miles/h [E]
The following examples are showing different ways we can combine the equation list we have to date. I will show a problem on a slide. I will show the solution to the problem on the next slide so that you can try it by yourself before seeing the answer. Current Equations: Δd = d 2 d 1 Δt = t 2 t 1 displacement time interval from instantaneous clock times Δd total = Δd1->2 + Δd2->3 total displacement if you were given individual displacements for movement but no reference point Δttotal = Δt 1->2 + Δt 1->2 D = Δd if no turning D = Δd1->2 + Δd2->3 v = D Δt speed total time interval if you were given individual time intervals for movement but no instantaneous clock times if turns once (more turns, more terms!) refer to the last handout you received in class about time and look at the last example...you would need to add each time interval once you had converted them all to either minutes or seconds. v = Δd Δt v = Δd Δt used if object is moving at a constant velocity used if object is moving at changing velocities
EX. Mr. Whitrow walks 120 m north in 4.15 minutes. He then walks 135 m south in 3.75 minutes. a) What is Mr. Whitrow's average velocity in km/h? b) What is Mr. Whitrow's average speed in m/s?
EX. walks north walks south total displacement no reference point given Mr. Whitrow walks 120 m north in 4.15 minutes. He then walks 135 m south in 3.75 minutes. a) What is Mr. Whitrow's average velocity in km/h? Δt 1 = 4.15 min x 1 h 60 min Δt 2 = 3.75 min x 1 h 60 min = 0.0691667 h = 0.0625 h Δd total = Δd1 + Δd2 = 0.120 km +(-0.135 km) = - 0.015 km [N] Δd 1 = 120 m [N] = 0.120 km [N] Δd 2 = 135 m [S] = - 135 m [N] = -0.135 km [N] Δt total = Δt1 + Δt2 = 0.0691667 h + 0.0625 h = 0.0940667 h v =? v = Δdtotal Δttotal = - 0.015 km 0.0940667 h v = - 0.15946 km/h = - 0.16 km/h [N] b) What is Mr. Whitrow's average speed in km/h? The cool thing is that I can use calculated values from part (a) to solve part (b). This will reduce the work. Δd 1 = 0.120 km [N] Δd 2 = -0.135 km [N] Δt total = 0.0940667 h v = D total = 0.255 km Δt total 0.0940667 h = 2.7108 km/h D= Δd1 + Δd2 = 0.120 km + -0.135 km = 0.120 km + 0.135 km = 0.255 km use absolute value signs to make these into scalar measures of length with no regard to direction = 2.7 km/h speed is a scalar so no direction
More examples: Jake is originally located 1.28 km [S] of Colonel Gray. After 25 minutes, Jake is located at 4.32 km [N] of Colonel Gray. What is Jake's average velocity in km/h? Mr. Whitrow drives at an average velocity of 13.6 m/s east for 38.5 minutes. If he stopped 18 km west of Colonel Gray, where did he start the trip?
More examples: Jake is originally located 1.28 km [S] of Colonel Gray. After 25 minutes, Jake is located at 4.32 km [N] of Colonel Gray. What is Jake's average velocity in km/h? d 1 = 1.28 km [S] of Colonel Gray = - 1.28 km [N] Colonel Gray d 2 = 4.32 km [N] of Colonel Gray is a reference point Δt = 25 min x 1h = 0.416667 h 60 min v =? Jake's displacement d 1 0 km d 2 where he started Colonel Gray where he stops Δd = d 2 - d 1 = 4.32 km - ( - 1.28 km) = 5.60 km [N] v = Δd = 5.60 km Δt 0.416667 h = 13.43999 km/h v = 13 km/h [N] Mr. Whitrow drives at an average velocity of 13.6 m/s east for 38.5 minutes. If he stopped 18 km west of Colonel Gray, where did he start the trip? v = 13.6 m/s [E] Δt = 38.5 min x 60 s = 2310 s 1 min v = Δd Δt Δd = v Δt d 2 = 18 km [W] = - 18 x 10 3 m [E] d 1 =? = (13.6 m)(2310 s) = 31416 m [E] s Δd = d 2 d 1 d 1 = d 2 Δd When subtracting big numbers, I prefer to put them into scientific notation and to the same power of ten. This allows me to count out the decimal places I can keep when subtracting. = 1.8 x 10 4 m 3.1416 x 10 4 m = 4.9416 x 10 4 m = 4.9 x 10 4 m [E]
EX. Mr. Cole leaves his house at 4:15 pm. His house is 0.26 km north of the school. He arrives at a store that is 4.89 km south of the school at 4:27 pm. What is his average velocity in km/h? Does this mean he drove at this velocity throughout the trip?
EX. Mr. Cole leaves his house at 4:15 pm. His house is 0.26 km north of the school. He arrives at a store that is 4.89 km south of the school at 4:27 pm. What is his average velocity in km/h? d 1 = 0.26 km [N] of school d 2 = 4.89 km [S] of school = - 4.89 km [N] t 1 = 4:15 pm t 2 = 4:27 pm Δt = t 2 - t 1 = 4:27pm - 4:15 pm = 12 min v =? Δt = 12 min x 1h = 0.20 h 60 min Δd = d 2 - d 1 = - 4.89 km - 0.26 km = - 5.15 km [N] v = Δd = - 5.15 km = - 25.75 km/h Δt 0.20 h v = - 26 km/h [N] Does this mean he drove at this velocity throughout the trip? This does NOT mean that he drove-26 km/h north for the entire trip. He could have slowed down, sped up, and/or changed directions several times during this trip.
Sam starts running at 3:25 pm at a position 980 m north of his house. He runs at an average velocity of 4.25 km/h south until he is 3.15 km south of his house. At what clock time did he stop and call the coach?
Sam starts running at 3:25 pm at a position 980 m north of his house. He runs at an average velocity of 4.25 km/h south until he is 3.15 km south of his house. At what clock time did he stop and call the coach? d 1 = 980 m [N] of HOUSE = - 0.980 Km [S] d 2 = 3.15 km [S] of HOUSE t 1 = 3:25 pm t 2 =? v = 4.25 km/h [S] Δd = d 2 - d 1 = 3.15 km - (-0.980 km) = 4.13 km [S] v = Δd Δt Δt = Δd = 4.13 km = 0.97176 h v 4.25 km/h Δt = 0.97176 h x 60 min 1 h = 58:306 min = 58 min...round to nearest minute to make clock times easier Δt = t 2 - t 1 t 2 = t 1 + Δt = 3:25 pm + 58 min = 4:23 pm It is one hour less two minutes than when he started.
Yes, you can combine steps to make more complicated equations. I will discuss what that looks like in class. At this point in the semester it is not expected that you would find that easy. However, as we get going you will develop the skill. And, hopefully, see how it can actually save you work. In grade 12 it is a necessity at times because you might have an unknown value cancel out.
(Really would benefit you to work out the examples first with your calculator. Do them once, twice, or even three times if needed before proceeding.) Homework on next page:
If preceeding questions went well, try these slightly more involved problems:
Attachments basic example constant velocity vs. average velocity distance and displacement