3. Solved Examples
1. When a comet is far from other objects, what can we say about its velocity? Its position? Its acceleration?
Solution:
Velocity: The velocity will be approximately constant (both magnitude and direction) according to Newton's first law.
Position: The position will change linearly with time, given the constant velocity.
Acceleration: The acceleration will be approximately zero, as there are negligible external forces acting on it.
2. The position of a coffee cup on a table as referenced by the corner of the room in which it sits is . How far is the cup from the corner? What is the unit vector pointing from the corner to the cup?
Solution:
Distance:
Unit vector:
3. Make a time-dependent position vector that describes the cup from the previous problem such that it starts in the location given, but slides at a constant rate of 1 m/s in the direction. Show that your position vector gives the right velocity by doing the math.
Solution:
Position vector:
Velocity vector: . The velocity vector is constant and equal to 1 m/s in the direction, as desired.
4. The velocity of a plane is given by . What is the speed of the plane? What is its direction of travel? If the x-direction is east and the y-direction is north, at what angle north of east is the plane flying?
Solution:
Speed:
Direction: north of east.
5. Give estimates of how fast you can run, how fast a car drives on the freeway and how fast jet planes fly in meters per second.
Solution:
Running: ~5-10 m/s
Freeway driving: ~25-40 m/s
Jet plane: ~200-300 m/s
6. Given a position function of , what is the velocity function? What is the velocity vector at t=3.0s?
Solution:
Velocity function:
Velocity at t=3.0s: m/s
7. What one-dimensional position function must an object have such that the object's position is numerically equal to its speed at all moments in time? In other words if it's at then it's also traveling at that moment. Solve by writing out an appropriate equation first.
Solution:
We want , and . So we have . Separating variables, we get . Integrating both sides, we get . Exponentiating both sides, we get , where is a constant. So .
8. Given a velocity of what must the units of the numbers '5' and '3' be? What will the displacement of the object described by this function be in the interval between t=1s and t=4s? If the object started at when t=1, where would it be at t=4s?
Solution:
Units: The '5' must have units of m/s, and the '3' must have units of m/s2 to make the equation dimensionally consistent.
Displacement:
m.
Position at t=4s: m.
9. What is the average velocity of a race car that does exactly 440 laps around a one mile circuit in 4.0 hours? What is the magnitude of the average velocity? What is the average speed of the race car?
Solution:
Average velocity: Since the car ends up where it started after 440 laps, the displacement is zero. Therefore, the average velocity is zero: .
Magnitude of average velocity: 0.
Average speed: The total distance traveled is 440 miles. The average speed is distance/time = miles per hour. Converting to meters per second: .
10. A cyclist races in a straight line with a speed function in meters per second given by , where a,b and c are constants. Define a function in GeoGebra of this form and adjust the constants so that after a long time the cyclist is going 13 m/s, and early on reaches a maximum speed of 20 m/s, 8 seconds from the start. What values of a,b and c accomplish this (with correct units)? How far will the cyclist have gone in 60s? What will the cyclist's speed be at t=10s? How long will it take to cover one kilometer? Please use GeoGebra to do this. At what time is the acceleration maximum? Minimum?
Solution:
This problem requires experimentation in GeoGebra. Here's a likely solution.
After a long time, , so .
Maximum speed of 20 m/s at t=8s suggests tuning a and b. Through trial and error: , work pretty well.
So
To find the distance after 60s, integrate from 0 to 60: . You can evaluate this numerically in GeoGebra (or other software) to get approximately 861.4 meters.
Speed at t=10s:
Time to cover one kilometer: Solve for t. Again, use numerical methods in GeoGebra. This gives approximately 69.4 seconds.
Acceleration: . So . To find the max/min of acceleration, we would take another derivative (the jerk!) and solve for zero (which is above the level of the students in this problem). Let's just evaluate the acceleration in GeoGebra to determine its max and min: . Graphing this on GeoGebra shows the maximum acceleration at about t=3.7s and minimum acceleration as t goes to infinity, approaching 0.
11. An ant runs along a piece of paper following a plot of the function . One unit on the paper is a centimeter. If the ant starts at the origin, how far will it have run when it gets to x=5.0 cm? If it takes 4.0s, what was its average speed? Average velocity? The magnitude of the average velocity? Use GeoGebra to do this problem.
Solution:
Arc length: The distance the ant runs is the arc length of the curve from x=0 to x=5. The arc length formula is . In this case, , so . You can evaluate this numerically in GeoGebra to get approximately 12.78 cm.
Average speed:
Average velocity: The displacement vector is (since ). So cm/s
Magnitude of average velocity:
12. Make a plot of the acceleration of a ball that is thrown upward at 20 m/s subject to gravitation alone (no drag). Assume upward is the +y direction (and downward negative y).
Solution:
The acceleration is constant and downward, so . The plot would be a horizontal line at -9.8 m/s2.
13. Make a plot of the same ball's velocity.
Solution:
The velocity is . The plot would be a line with a y-intercept of 20 m/s and a slope of -9.8 m/s2.
14. Make a plot of the same ball's position.
Solution:
The position is . The plot would be a parabola opening downward, with a y-intercept of 0.
15. At what time will the ball be back at its starting position?
Solution:
We want to find when . From the previous problem, . This occurs at t=0 (the starting time) and when , which means .
16. At what time will the same ball be at max height?
Solution:
The ball will be at max height when . From problem 13, . Solving for t, we get .
17. What will the velocity in the y direction be at max height?
Solution:
The velocity in the y direction will be 0 at max height.
18. Do any of these answers change if the ball additionally has a horizontal component to its initial velocity?
Solution:
No. The horizontal component of the initial velocity does not affect the vertical motion of the ball (assuming no air resistance). The time to reach max height, the max height itself, and the time to return to the starting position depend only on the initial vertical velocity and the acceleration due to gravity. The acceleration is constant, so it doesn't depend on the starting horizontal velocity.
19. Name these variables:
Solution:
: Displacement
: Position
: Change in velocity
: Velocity
: Speed (magnitude of velocity)
: Acceleration
: Magnitude of the acceleration
: Average velocity
: Time
: x-component of the velocity
20. A car driving at 24m/s veers to the left to avoid a deer in the road. The maneuver takes 2.0s and the direction of travel is altered by 30 degrees. (Don't worry about the fact that they will later veer back into their own lane.) What is the average acceleration during the constant speed maneuver? Do this in accordance with the example in the chapter.
Solution:
First, we need to recognize that only the direction of the velocity is changing and not the speed. This means that if we think of and take the derivative using the chain rule that we get the first term to be zero (due to constant magnitude) and so . The unit vector is only direction. It's change indicates (in radians per second) how quickly the direction is changing. Here this is 30 degrees, or over a period of 2.0s. So the acceleration is a vector with magnitude. The direction, as we will study more in a future chapter is in the lateral direction perpendicular to the velocity.