Chapter 2: Motion in One Dimension

When velocity is negative, is acceleration always negative?

No, acceleration is the slope of the velocity-time graph, it can be positive or negative and does not depend on is velocity •When would this be the case, negative to positive velocity, with positive constant acceleration?

An object starts at rest, undergoes positive, constant acceleration for 10 s, then continues on with constant positive velocity, Which x-t graph correctly describes this situation?

An x-t graph would look like a positive parabolic trajectory from 0 s to 10 s, then it would have a constant slope the rest of the way

An x-t graph during a 5 s interval has a positive constant slope until 2 s, then it is a straight line for a few seconds, then ends with a negative constant slope until 5 s but is much steeper than first, What would the v-t graph look like?

Each section has a constant velocity, the first part would be positive, over the x-axis, the second part would be equal to 0 so right on the x-axis, and the third was is negative velocity so would be under the x-axis at furthest because its x-t graph was steeper

The velocity at the 2 second point of the x-t graph is?

Need to get the slope, rise over run, It goes up 5 m after 2 s, so 5/2 gives us a slope of 2.5 m/s

Looking at the x-t graph, It starts off as a straight line, after half of the time interval is elapsed, the position goes down parabolically and goes into a straight line again at the end, What is the best way to explain this?

The object doesn't move at first, then it moves backwards(or down) and then finally stops

Which is the best interpretation of a graph of x-t going diagonally to the right at the same slope throughout?

The object is moving at a constant velocity

Looking at an a vs. t graph

•If we rearrange the equations, Δv = aΔt -Say we have initial values at 0, and go to time t' and acceleration a', The are of the a' x t' box would be equal to the change in velocity, this is how we go in the reverse direction, sum of squares •So from Δx to Δv to a, we use the slope concept(rise over run to get triangle slope), and to go from a to Δv to Δx, we use the area concept

What we actually measured was the water bottle's Average Velocity

•Instantaneous Velocity is the velocity at an exact moment in time, not over a long time interval •vinst = limit(Δt → 0): Δx/Δt -This is never really measured in real life, the smaller and smaller the time interval is, the vavg gets closer to vinst

Vector

•Numbers with magnitude and direction, ex. displacement, xf - xi or Δx, Δ meaning "change in..." -A vector will always be ± some number -If we were to walk 1 meter from a spot and then walk back to it, our displacement would be 0 m, but distance would be 2 m, distance is not a vector unlike displacement, which has both magnitude(#) and direction(±)

Variables we've discussed so far

•Position(x) •Displacement(Δx) •time(t), "clock reading" at an exact time, which is different than the vector, time interval(Δt or tf - ti) -Time interval is always positive, unless going back in time like in time travel, which is mathematically possible in quantum mechanics because of the uncertainty principle, but not for the scope of our class

Position

•The number on a ruler closest to some object of interest -The object of interest was his water bottle, The ruler was a 2 meter stick, he placed it at the 0.5 m or 50 cm mark, so that is its position, x = 50 cm, You always need a variable(x), a comparater(=), a value(50), and a unit(cm) -He then moved the water bottle to the other side of the 2 meter stick, now x = 1.5 m or 150 cm -He then moved it behind the 2 meter stick, now x = -0.5 m or -50 cm

Uniform Velocity and Uniform Acceleration

•Uniform Velocity -In the motion diagram, the marks are evenly spaced apart, they are travelling at the same speed(magnitude) and same direction every units of time for the motion diagram -Every time interval, the same distance is covered -x vs. t graph: It is a line with the same slope throughout, if it were faster it would have a steeper slope, and if it were slower it would have less steep of a slope -v vs. t graph: It would be a straight line across because the velocity is constant, if it were faster it would be higher up and if it were slower it would be more down, but still a straight line -a vs. t graph: Acceleration would be 0 for all 3 cases, no matter how large or small of its velocity, It is its rate of change of v, since v is not changing then a = 0 •Uniform Acceleration(toy car with fan example) -a vs. t graph: It would be a straight line across, if it has a larger acceleration it would be higher up, and if it were lower, it would be lower down, but still a straight line because its acceleration is constant, a = Δv/Δt, in this case ainst = aavg -v vs. t graph: It is a line with the same slope throughout, if it had a larger acceleration, then its slope would be steeper(larger), it has a big velocity change over a smaller time interval compared to it with a lower acceleration, The slope gives us information(rise over run, draw a triangle), If the velocity was constant, vinst would be equal to vavg, they are in the same ratio -x vs. t graph: It would have a parabolic trajectory with its slope getting higher each unit of time at the same rate, The slope is getting bigger, Tangential lines are drawn and their slopes are increasing each time, remember v = rise over run of this x vs. t graph, and An object's displacement, Δx = Δv X Δt, the area under the v-t graph

Over the Weekend

•We watched a toy car with a fan propelling it forward, Was this uniform motion?, It started slow and sped up •What would x, v, and a vs. t graphs look like for this motion?

Instantaneous Acceleration

•a = limit(t → 0): Δv/Δt -Acceleration is the rate of change of velocity(just like velocity is the rate of change of displacement) -If we were to graph from the same table above, a vs. t, it would be x(a) = 0 at any y(t), because the velocity did not change at all, time is changing, but since the rate of velocity isn't then it will always equal zero -We could have a negative value for acceleration and velocity if it were travelling in the other direction

Summary of Chapter 2

•v = Δx/Δt •vinst = limit(Δt→0): Δx/Δt •a = Δv/Δt •ainst = limit(Δt→0): Δv/Δt -These involve looking at the graphs and converting them(slope and area concepts) •Special Case Equations with Uniform Acceleration(Cardinal Equations) 1.) Δx = v0•t + 1/2 a•t² 2.) v = v0 + a•t 3.) v² = v0² + 2a•Δx -v0 refers to initial velocity, v refers to final velocity(or any velocity your'e looking at) -These can be rearranged to find other values, and will help for word problems involving these variables

HW #6.) A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42 m/s due north, After a reaction time of 0.800 s the policeman begins to pursue the speeder with an acceleration of 5.00 m/s², Including the reaction time, how long does it take for the police cart to catch up with the speeder?

•vp0 = 18 m/s, vc0 = 42 m/s •ap = 5 m/s², ac = 0 m/s² •Reaction time = 0.800 s before police accelerates •We need to calculate the distance the cars travel during that reaction time, because x0(once cop accelerates) for both will not be 0 m, we can solve for them with: Δx = vΔt •Δxp = vΔt = 18 m/s•0.8 s = 14.4 m = xp0 •Δxc = vΔt = 42 m/s•0.8 s = 33.6 m = xc0 •When t = 0, is when the police car starts accelerating •Using this equation(rearranged from cardinal equation): x = x0 + v0t + 1/2 at², we can write equations for both •xp = 14.4 m + 18 m/s•t + 1/2(5 m/s²)t² •xc = 33.6 m + 42 m/s•t •When crossing, they will have the same position, so xp = xc, We can set them equal to each other and solve for t •14.4 m + 18 m/s•t + 1/2(5 m/s²)t² = 33.6 m + 42 m/s•t •2.5 t² - 24 t - 19.2 = 0, We solve using the Quadratic Formula •We get t = 10.34 s or -0.74 s(which is not possible) •So it takes the police car 10.34 seconds to catch up to the speeder, once it starts accelerating, which plus RT = 11.14 s when it first passed the police car

Problem 2.2: Porsche - Spotting a Police Car

A Porsche is travelling at 100 km/hr but sees a police car, so slows down to 80 km/hr during a displacement of 88 m, Assume constant acceleration, What is the car's acceleration? What is the time(t) required for a given decrease in speed? Start with the givens: •vi = 100 km/hr, vf = 80 km/hr, Δx = 88 m •Given these variables, we can use the following cardinal equation -vf² = vi² + 2a•Δx -We can rearrange it to be a = (vf² - vi² / 2Δx) -We need to convert units into m/s, 100 km/hr(1 hr/60 min)(1 min/60 sec)(1000 m/1 km) = 27.8 m/s, and 80 km/hr(1 hr/60 min)(1 min/60 sec)(1000 m/1 km) = 22.2 m/s -a = (22.2 - 27.8) / 2•88 = -1.59 m/s² -The Porsche accelerated at -1.59 m/s²(opposite sign of velocity, so it is speeding down obviously) •How much time did it take? -Use the following cardinal equation: v = v0 + a•t(easier to do than the other one which makes you do quadratic formula) -Rearrange it into t = (vf - vi)/a -t = (22.2-27.8)/(-1.59) = 3.52 sec -So it took the Porsche 3.52 sec to slow down to its new, legal speed, We don't care about sig figs, just keep it to hundredths digit

Speeds(+) vs. Velocity(+ or -)

A ball is thrown into the air at 10 m/s, v0 is fastest right at the point of release, it then will slow because of gravity until it is 0 m/s, then it will fall down and increase speed with gravity •v-t graph, It would start at +10 m/s at t = 0, it would go down at a constant slope, when it crosses the x-axis, is at the balls peak, highest point, v = 0 m/s, this is the turning point, it then has negative velocity and increases in magnitude •s-t graph, It would start at 10 m/s then go down with a constant slope until the x-axis, where s = 0 m/s, this is the peak or turning point, it then would go back up because the speed is increasing(gravity accelerates it), and is positive because speed does not indicate direction, only its magnitude -s = |v| -s = |+10 m/s| = 10 m/s -s = |-10 m/s| = 10 m/s

Problem 2.3: Police Car Chase

A car is travelling at a velocity of 90 km/hr, well over the speed limit, When it passes a parked police officer, the police car then accelerates at 3.5 m/s², How long until the police catches up to the speeding car? What distance did that car travel? Start with the givens? •vcar = 90 km/hr, acop = 3.5 m/s² •Δxcop = v0•t + 1/2 acop•t², since xicop = 0 and v0 = 0, then we can rearrange the equation into xfcop = 1/2 acop•t² •Δxcop = Δxcar, they both travel the same distance until they meet again so can use the cars, We can substitute vcar•t for Δxcar, since we know these values as well •vcar•t = 1/2 acop•t², this can rearrange into t = 2 vcar/acop, now we just need to do conversions and plug in -90 km/hr needs to be converted into m/s, 90 km/hr(1 hr/60 min)(1 min/60 sec)(1000 m/1 km) = 25 m/s -Plug them in, t = 2(25 m/s)/(3.5 m/s²) = 14.3 sec -It took the police car 14.3 seconds to catch up to the speeding driver •What is the total displacement of the speeding car? -vcar = Δxcar/tcar, we can rearrange into Δxcar = vcar•tcar, since we have both of those values, then calculate -25 m/s•14.3 s = 357 m -The speeding car had a total displacement, when passing the police car until it caught up to them again, of 357 meters

Uniform Acceleration Demo

A good example of this is a vertical free fall •Using a device that precisely measures the time of the free ball of a marble, wind resistance is not very effective, marble slices right through it •Drawing a motion map, it will start at H of 1.5 m or 150 cm, the marks will get more spaced out each time unit(t1.t2..t3...t4....5f, etc.), until it hits the floor at x = 0 Let's predict the time of the fall(T) •a = 9.8 m/s², H = 1.5 m •We have our three cardinal equations for uniform acceleration 1.) Δx = vi•t + 1/2 a•t² 2.) v = vi + a•t 3.) v² = vi² + 2a•Δx •We can infer vi = 0 m/s(before it is dropped), Δx is -1.5 m, and g or a = 9.8 m/s² •The only formula that would make sense is Δx = vi•t + 1/2 a•t² •-1.5 m = 0 m/s•t + 1/2 9.8 m/s²•T² •-1.5 m = 4.9 m/s²•T² •T² = (-1.5 m)/(4.9 m/s²) •T² = -0.306 s², the square root of this gives us T = 0.553 s In the actual experiment •Tactual = 0.549 s, Tpredicted = 0.553 s •%different = |predicted - actual| / actual X 100% •|0.553 - 0.549| / 0.549 X 100% = 0.729 % error, less than 1% is very accurate

Motion Diagram

Same procedures as above(right to left is positive direction) •If speeding up: x..........x.....x...x..x •If slowing down: x..x...x.....x..........x -The space between the marks(Δx) gets bigger when speeding up, We could get each of their positions with a very large ruler

Which is acceleration most negative in this v-t graph?

Section X to Z has the steepest negative slope, so is the answer

Looking at a v-t elevator graph, how far does it travel in the first 3 seconds?

Using the area under the curve of the triangle, 1/2•4 m/s•3 s = 6.0 m, Displacement is known as area trapped under the v-t graph

What was the car's acceleration at the 90 second mark using this v-t graph?

We can get any triangle under the time from 60 to 120 seconds because it has a constant acceleration, 20 m/s - 10 m/s / 90 s - 60 s gives us a slope of 0.33 m/s²

Problem 2.1: Out of Gas

Your pickup truck travels 8.4 km at 70 km/hr, You then run out of gas and walk for 30 min to a gas station 2 km away, Assume constant velocity, What is the overall displacement?, What is the total time it takes?, Also what is the average velocity? Start with the givens: •Δxpickup = 8.4 km, Δtpickup = ?, vpickup = 70 km/hr -We can rearrange v = Δx/Δt into Δt = Δx/v -Δtpikcup = 8.4 km/70 km/hr = 0.12 hr -So Δtpickup = 0.12 hr •Δxwalk = 2 km, Δtwalk = 30 min or 0.5 hr, vwalk = ? -v = Δx/Δt -vwalk = 2 km/0.5 hr = 4 km/hr -So vwalk = 4 km/hr •What is the overall displacement? -Δxtotal = Δxpickup + Δxwalk -Δxtotal = 8.4 km + 2 km = 10.4 km -So the overall displacement is 10.4 km •What is the total time? -Δttotal = Δtpickup + Δtwalk -Δtotal = 0.12 hr + 0.5 hr = 0.62 hr -So the total time it took was 0.62 hr or 37.2 min •We can also calculate the average velocity -vtotal = Δxtotal/Δttotal -vtotal = 10.4 km/0.62 hr = 16.8 km/hr -The average total velocity(same as speed, because no change in direction) was 16.8 km/hr

Scalar

•A number that only has magnitude, not direction •Distance and speed(s), which are different than displacement(Δx) and velocity(v), which are vectors •The frequency of sound is also a scalar, Absolute Temperature is as well because they only indicate magnitude, but no direction

Velocity, v(m/s)

•Also a vector, they are sometimes written with a "→" over their symbol to indicate so •Δx/Δt(m/s) -If the water bottle went from 100 cm mark to 0 cm mark in 2 seconds, Δx = -100 cm, Δt = 2 s, so -100 cm/2 s = -50 cm/s or -0.5 m/s, Without the + or - sign, it just indicates speed(which is not a vector), it has no direction(always positive)

HW #4.) Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff, The height of the cliff is 6.00 m, The stones are thrown with the same speed of 9.00 m/s, Find the location (above the base of the cliff) of the point where the stones cross paths

•Ball down(1), v0 = -9 m/s, x0 = 6 m •Ball up(2), v0 = +9 m/s, x0 = 0 m •Cardinal Equation: Δx = v0•t + 1/2 a•t² or x - x0 = v0•t + 1/2 a•t² •We are looking for xf, so can rearrange it to be: x = x0 + v0•t + 1/2 a•t² •Do equation for both balls •x1 = 6 m - 9t + 1/2 g•t² •x2 = 0 m + 9t + 1/2 g•t² •Since they will have the same final position, we can set them equal to each other, x1 = x2: 6 - 9t + 1/2 g•t² = 9t + 1/2 g•t² •1/2 g•t² is on both sides so cancels out, It simplifies to: 6 = 18t, so t = 0.33 sec •You could then plug it into either x1 or x2 equations because they have the same final position •x1 = 9 m/s•t + 1/2 g•t² •x1 = 9 m/s(1/3 s) + (1/2)(-9.8 m/s²)•(1/3 s)² •x1 = 3 m - 0.54 m •x1 and x2 = 2.46 m

Constant Velocity with Metronome counting off seconds

•If he were to walk across the board at a constant velocity, starting at x = 0, t = 0, and the positive direction being to the left, and would mark an x every second the Metronome goes off -There would be marks placed evenly away from each other at the same distance, there was no change in velocity -Note that if there is no change in position, then velocity would also be considered constant because it would be 0 m/s, there is no Δx, only Δt, but it doesn't matter because 0 is in the numerator •These marks placed across the board are called a Motion Diagram -x.....x.....x.....x

Basic Graphs of Changes in Displacement(x), Velocity(v), and Acceleration(a) vs. Time(t)

•x vs t. graph -line a starting at 0 and going up a slope of 10 m -line b starting at 1 and going up a slope of 10 m -line c starting at 2 and going down a slope of -10 m -They are all at a constant velocity motion, AKA Uniform Motion -They do not depend on x because Δx is constant, They are all at an even rate •v vs. t graph -lines a and b would look the same and be straight lines because of being constant, they have different initial and final positions, but there is no differences between their constant instantanenous velocities, they would both be x = 10 m/s -line c would be a straight line as well because of being at a constant velocity, but it would be at x = -10 m/s, because it is travelling at the same speed, but in the opposite direction, the slope on the x vs. t graph is going the opposite way as the other two •a vs. t graph -lines a, b, and c would all be at x = 0 m/s², because there are no changes in velocities for either of them, they are at constant velocities the entire time, so no acceleration is taking place So as we can tell, position-time graphs rarely ever look like v-t graphs and so on of the same data, they are rates of changes of each other, derivatives, So if a graph of a slope of 1, the next graph would be x = 1, which now as a slope of 0, so the next graph would be x = 0 and so on, likewise, they are the area under the curves to get the graph before it, i.e., area under the v-t graph is the total displacement

We can make a position and time table

•x(m), t(s) -0, 0 -10, 1 -20, 2 -30, 3 •We can graph it, vavg = Δx/Δt -We can take two points and draw a triangle to see the side lengths and the difference between the points, 30 m - 10 m and 3 s - 1 s, Δx = 20 m and Δt = 2 s, so vavg = 10 m/s -We can take a different triangle, now have 30 m - 20 m and 3 s - 2 s, Δx = 10 m and Δt = 1 s, so vavg is still = 10 m/s -We can even take a super tiny triangle and it would still have the same vavg, that is because velocity is always in a ratio of two numbers -If we were to graph this velocity an an instantaneous time, v(cm/s) vs. t(s) graph, it would be x(vinst) = 10 at any y(t), because it was traveling at a constant velocity

Displacement

•xfinal - xinitial or xf - xi -He placed the water bottle at 1 m(xi), then at 0.5 m(xf), so the displacement or Δx = -0.5 m or -50 cm -He moved the ruler, but the water bottle still moved the same, so its displacement was still -0.5 m or -50 cm, xf and xi just changed but didn't affect Δx, xf now = 150 cm and xi now = 200 cm, giving us Δx = -50 cm, It doesn't depend on how you translate the ruler, displacement doesn't change, it is invariant to the rulers movement -If it displaced the other way, then Δx would be +50 cm, going from 200 to 250 cm instead, It traveled the same distance, but had a different displacement, The + or - sign gives you the direction of the displacement

HW #2.) A ball is thrown vertically upward, which is the positive direction, A little later it returns to its point of release, The ball is in the air for a total time of 8.0 s, What is its initial velocity? Neglect air resistance

•Δ(total) = 8 sec •Δ(peakpoint to hand) = 4 sec •Cardinal equation: v = v0 + aΔt •Since v0 = 0 m/s, and a = g = -9.8 m/s², We change it to: v = gΔt •v = -9.8 m/s²•4 s •v = -39.2 m/s •The velocity of the ball returning to hand is -39.2 m/s, By symmetry, the (initial)velocity of the toss is +39.2 m/s •Alternatively: -v(final) = v(initial) + gt(8 s), v = speed