Съдържание

Условие Отговори Решение

USAPhO 2004, Четвъртфинал — Задача 1. Acceleration vs. Time

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2004

AAPT AIP United States Physics Team 2004 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie
Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Average Earth-Sun Distance	1 Astronomical Unit (A.U.) $= 1.5\times10^{11}\ \mathrm{m}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Moment of Inertia about Center of Mass – Uniform Object (will not be provided on the second screening exam)

Disk	$^1/_2\,MR^2$
Sphere	$^2/_5\,MR^2$
Rod	$^1/_{12}\,ML^2$

Solutions to Problems

Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 4.a., if a student wrote down $\omega = (Mv\alpha L)/(2mL^2)$ , they should get 6 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

Graph of acceleration a versus time t: a = a0 from 0 to t0, a = 0 from t0 to 2t0, a = –a0 after 2t0. Axis labels a, t; tick labels a0, –a0, t0, 2t0; title Acceleration vs. Time. — Acceleration vs. Time

An object starts at rest at the origin at $t = 0$ and accelerates according to the graph shown above. The motion is one-dimensional.

a. Sketch a graph of velocity versus time. Show and clearly label all extreme values, discontinuities, and zero crossings. [10 т.]

b. At what time (expressed in terms of $t_0$ ) does the object return to the origin? [15 т.]

Отговори

Решение

1. a. The graph should look like the following.

Both axes are labeled [1]

Shows $v = 0$ at $t = 0$ [1]

Straight line with positive slope from $t = 0$ to $t = t_0$ [1]

Shows discontinuity in slope at $t = t_0$ [1]

Labels maximum velocity at $v = a_0t_0$ [1]

Shows constant velocity from $t = t_0$ to $t = 2t_0$ [1]

Shows discontinuity in slope at $t = 2t_0$ [1]

Straight line with negative slope after $t = 2t_0$ [1]

Shows $v = 0$ at $t = 3t_0$ [1]

No spurious features shown [1]

(There are at least two methods that can be used to find the time at which the object returns to the origin. These are detailed below. Students may be awarded points from only one method.)

b. Method I: The distance traveled is given by the area under the velocity vs. time curve. [5]

Continue the curve until the total area is zero, i.e., the area below the $t$ -axis is equal to the area above it. Find the value of $t$ at which the object returns to the origin either by construction or geometrical formula. The construction method follows.

The area above the time axis is $d_+ = 2a_0t_0^{\,2}$ . [4]

To make the area below the axis equal, extend the graph $2t_0$ beyond the zero crossing at $3t_0$ . [4]

and $t = 2t_0 + 3t_0 = 5t_0$ [2]

(Students should be given full credit for any valid geometrical argument.)

Method II: For the interval $t = 0$ to $t = t_0$ , $a = +a_0$ and $v_0 = 0$ . The kinematic equations for motion with constant acceleration yield

At time $t = t_0$ , $v = a_0t_0$ [2]

$x = \frac{1}{2}a_0t_0^{\,2}$ . [2]

These become the initial velocity and displacement for the next interval $t = t_0$ to $t = 2t_0$ . During this interval, $a = 0$ and $v_0 = a_0t_0$ .

At time $t = 2t_0$ , $v = a_0t_0$ [2]

$x = x_0 + v_0(2t_0 - t_0) = \frac{1}{2}a_0t_0^{\,2} + (a_0t_0)t_0 = \frac{3}{2}a_0t_0^{\,2}$ . [2]

These become the initial velocity and displacement for the next interval $t > 2t_0$ . During this interval, $a = -a_0$ and $v_0 = a_0t_0$ .

For $t > 2t_0$ , $x = x_0 + v_0(t - 2t_0) - \frac{1}{2}a_0(t - 2t_0)^2 = \frac{3}{2}a_0t_0^{\,2} + a_0t_0(t - 2t_0) - \frac{1}{2}a_0(t - 2t_0)^2$ [4]

$x = \frac{1}{2}a_0[3t_0 - (t - 2t_0)][t_0 + (t - 2t_0)] = \frac{1}{2}a_0(5t_0 - t)(t - t_0)$

It returns to the origin when $x = 0$ . This happens at $t = 5t_0$ . [3]

The other solution is eliminated because the equation is only valid in the time interval $t > 2t_0$ .

Velocity v versus time t: v rises linearly from 0 to a0t0 at t0, stays constant until 2t0, then decreases linearly, crossing zero at 3t0 and continuing negative past 4t0. Labels v, t, a0t0, t0, 2t0, 3t0, 4t0. — Velocity vs. Time

Construction on the velocity graph: the triangle above the axis (area under the curve up to 2t0) is moved by a curved arrow; shaded area above the axis equals shaded triangle below the axis. Dimension arrows 2t0 and 3t0 along the axis and 2t0 beyond the zero crossing; axes v and t.

Оригинал в Архива: Quarterfinal 2004.pdf · официални решения: Quarterfinal 2004 Solutions.pdf