QUIZ 2 (ANSWERS) |
CHAPTER 2 |
EXERCISE/PROBLEMS |
Problems labeled "mup" discussed in class. |
mup1 . Think of two cars starting in opposite directions 200
m apart.| v1|*t + | v2| *t = 200 . Assuming v1 > v2, angle with vertical = 180 - [360 - | v1 |*360*/(| v1 | + | v2 |)] = | v1 |*360*/(| v1 | + | v2 |)] - 180 = 360*{| v1 |/(| v1 | + | v2 |) - 1/2}. When |v1 | = |v2|, angle = 0. |
mup2. (a) and (b) The final version of this problem includes the
distance from the starting point . You can find the clockwise
distance for simplicity--find the total dance traveled,
then divide by 200 m for the number of revolutions. Say you
get 22.654. Find the product of the fractional part 0.654 and
circumference . In the current example multiply the total time for
the fast car in part (a) by speed 6.20 m/s. This will give
you the total distance fast car travels. Then divide by 200
m to get the number of revolutions--between 8.00 and 9.00.
Subtract 8.00 to get the fractional part f. Find product f*200 m = the number of meters from starting point clockwise. Note final answers are two sig. figs. |
10. The slope of the tangent at each time t equals the instantaneous velocity. Note: The velocity is constant during time intervals in which the x vs t curve is a straight line segment, whether flat or inclined. |
9. See section 2.6 for an illustration of
displacement as the area under the velocity curve. When you
divide that area by the time interval, you get the average velocity.
The average speed on the other hand is total distance/ total
time. |
12. The slope of the tangent at each time t equals the
instantaneous acceleration.. Note: The acceleration is constant
during time intervals in which the v vs t curve is a straight line
segment, whether flat or inclined. Average acceleration between any
two times = (change in v) /(change in t). |
16. The signs of average accelerations are (a) negative (b) negative (c) negative. Find magnitudes. |
mup3. You can plot v vs t from the derivative of x vs
t. For x vs t, the slope of the tangent at each time t equals
the instantaneous velocity. Note: The velocity is constant during
time intervals in which the x vs t curve is a straight line segment. There are 5 intervals: [0,5]: The slope increases from zero to a value equal to the slope of the line for interval [5,15] [5,15]: The slope is constant and positive [15,25]: The slope decreases to zero at t = 20 seconds and from zero decreases to a value equal to the slope of the line for interval [25,35]. [25,35]: The slope is constant and negative. [35, 40]: The slope increases to zero. You can plot a vs t from the derivative of v vs t. For v vs t , the slope of the tangent at each time t equals the instantaneous acceleration. Note: The acceleration is constant during time intervals in which the v vs t curve is a straight line segment, whether flat or inclined. There are 5 intervals: [0,5]: v increases linearly from zero; the slope is positive and constant [5,15]: v is constant; the slope is zero. [15,25]: v decreases linearly; the slope is negative and constant [25,35]: v is constant; the slope is zero. [35, 40]: v increases linearly to zero; the slope is positive and constant |
30. Acceleration is constant and negative. To get the total displacement find the sum of the areas under the curve; the areas are negative for t > 6.0 seconds. To get the total distance find the sum of the absolute values of the areas under the curve. |
31.. The accelerations are just the slopes of the line segments. To get the total displacement find the sum of the areas under the curve; the areas are never negative. The total distance in this case equals the total displacement since all motions are positive. |
mup4. The sketches of "a vs t" , "v vs t" and "x vs t" graphs look surprisingly like those of problem 19 . |
32. Two vehicles moving in the same direction. Indeed the graphs are faintly similar to those of #70 if the passenger train and freight train collide. The first crossing represents that collision , the smallest root of the quadratic equation like that described below. This problem's situation is different though: a car with constant velocity passes an accelerating car; the accelerating car later passes the car that has constant velocity. |
70. Solve 200 + 15*t = 25*t - (1/2)*(0.1)t^{2}. The left hand side of the equation represents the freight train's position, the right the passenger train. If they collide, compare the passenger train's and freight train's instantaneous velocity at the collision point. Explain the meaning of the velocity difference. |
72.. The cars' relative speed is 20 m/s. (See section 3.5) |
73..Another case of finding the intersection of two graphs: D + (1/2)*a'*t^{2} = (1/2)*a*t^{2} , where "a' " represent the truck's acceleration, "a " the car's acceleration, D the unknown initial difference in distance, and 40.0 m = (1/2)*a'*t^{2} ; also compare the truck's and car's instantaneous velocities as needed. |
77. The exercise belongs to a problem class defined by
73 and 32: The unifying theme is two objects moving
relative to each other. It demonstrates something
important for all motion problems: Identify the motion
of a specific point on the each extended object. For ease of
computations, 1-D motion uses point
idealizations. For this problem, the key points are
car's front
and truck's rear, all computations based on
distances between these two points which you can deduce from car's and truck's physical lengths. Lets' write the equation for car's front displacement after time t relative to the origin (x = 0): (20.0 m/s)*t + (1/2)*a*t^{2}, where the car's acceleration is given. Let's write the equation for the truck's rear displacement relative to zero (x = 0) after the same time t: 24.0 m + (20.0 m/s)*t. [staff notes: edit below~] We are given the displacement of car front = displacement of truck rear + 26.0 m + 21.0 m + 4.5 m = displacement of truck front + 51.5 m. In other words, (20.0 m/s)*t + (1/2)*a*t^{2} = 24.0 m + 51.5 m + (20.0 m/s)*t or (20.0 m/s)*t + (1/2)*a*t^{2} = 75.5 m + (20.0 m/s)*t. Solving this you can answer all questions. |