Straight Line Motion Revisited Homework Answers: Understanding Velocity vs Time Curves

To find the observable consequences of this curvature we need to take another step. When spacetime is flat freely moving objects obey Newton's first law i.e. they move in a straight line at constant speed. When spacetime is curved freely moving objects obey a different equation called the geodesic equation:

straight line motion revisited homework answers

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But all this is a bit abstract, and I suspect you're after a more intuitive feel for how does spacetime curvature cause straight lines not to be straight? Well the most obvious definition of a straight line is the path of a light ray, because we all learn in school that light travels in straight lines. And the obvious demonstration of this is the bending of light rays by the Sun.

If we shine a light ray so that it just grazes the Sun's surface then that light doesn't travel in a straight line. We can calculate the deflection using the geodesic equation and the values of the Cristoffel symbols, and we find the light ray is bent by about $1.75$ arcseconds. But this is a tiny, tiny amount. $1.75$ arcseconds is about the angle subtended by a baseball at a distance of $9$ kilometers - you couldn't even see a baseball $9$ km away!

How can you measure the horizontal and vertical velocities of a projectile? Vernier's Logger Pro can import video of a projectile. From the video, you can produce graphs and calculations of pretty much any quantity you want. Experimentally verify the answers to the AP-style problem above. Take video of two balls, perhaps launched with a Pasco projectile launcher so they are guaranteed to have the same initial speed. Launch one ball straight up, the other at an angle. Import the video to Logger Pro. Then check to see whether the speed of each ball is in fact the same at a given height.

The percentages, when converted to decimal values 0.15, 0.28, and 0.33,are the slopes of the straight lines which form the graph of thetax forthe corresponding tax brackets. The tax graph is what's called apolygonal line, i.e., it's made up of several straight line segmentsof different slopes. The first line starts at the point (0,0) and headsupward with slope 0.15 (i.e., it goes upward 15 for every increase of100 in the $x$-direction), until it reaches the point above $x=26050$.Then the graph "bends upward,'' i.e., the slope changes to 0.28. Asthe horizontal coordinate goes from $x=26050$ to $x=67200$, the linegoes upward 28 for each 100 in the $x$-direction. At $x=67200$ theline turns upward again and continues with slope 0.33. Seefigure 1.1.1.$\square$

The most familiar form of the equation of a straight line is:$y=mx+b$. Here $m$ is the slope of the line: if you increase $x$ by1, the equation tells you that you have to increase $y$ by $m$. Ifyou increase $x$ by $\Delta x$, then $y$ increases by $\Deltay=m\Delta x$. The number $b$ is called the y-intercept, becauseit is where the line crosses the $y$-axis. If you know two points ona line, the formula $m=(y_2-y_1)/ (x_2-x_1)$ gives you the slope.Once you know a point and the slope, then the $y$-intercept can befound by substituting the coordinates of either point in the equation:$y_1=mx_1+b$, i.e., $b=y_1-mx_1$. Alternatively, one can use the"point-slope'' form of the equation of a straight line: start with$(y-y_1)/(x-x_1)=m$ and then multiply to get $(y-y_1)=m(x-x_1)$, thepoint-slope form. Of course, this may be further manipulated to get$y=mx-mx_1+y_1$, which is essentially the "$mx+b$'' form.

The graph of $y$ versus $t$ is a straight line because you are travelingat constant speed. The line passes through the two points $(1,110)$ and$(1.5,85)$, so its slope is $m=(85-110)/(1.5-1)=-50$. The meaning of theslope is that you are traveling at 50 mph; $m$ is negative because you aretraveling toward Seattle, i.e., your distance $y$ is decreasing. The word "velocity'' is often used for $m=-50$, when we wantto indicate direction, while the word "speed'' refers to the magnitude(absolute value) of velocity, which is 50 mph. To find the equation of theline, we use the point-slope formula:$$ y-110\over t-1=-50,\qquad\hboxso that\qquad y=-50(t-1)+110=-50t+160.$$ The meaning of the $y$-intercept 160 is that when $t=0$ (when youstarted the trip) you were 160 miles from Seattle. To find the$t$-intercept, set $0=-50t+160$, so that $t=160/50=3.2$. The meaningof the $t$-intercept is the duration of your trip, from the startuntil you arrive in Seattle.After traveling 3 hours and 12 minutes, your distance $y$ from Seattle will be0. $\square$

The curve \(y(x)\) we are looking for minimizes the functional\[I(y)= \int_0^1ds = \int_0^1(1+(y')^2)^1/2 \; \rm dx\mboxsubjectto boundary conditionsy(0)=0,y(1)=1\] which means that \(I(y)\) has an extremum at \(y(x)\). It seems obvious that the solution is \(y(x)=x\), the straight line joining \((0,0)\) and \((1,1)\), but how do we prove this?

The constant \(\alpha\) is determined implicitly by the remaining boundary condition \(y(h)=a\). The equation of the cycloid is often given in the following parametric form (which can be obtained from the substitution in the integral) \[\beginalignedx(\theta) & = & \frac\alpha2(1-\cos2\theta) \\y(\theta) & = & \frac\alpha2(2\theta-\sin2\theta) \endaligned\] and can be constructed by following the locus of the initial point of contact when a circle of radius \(\alpha/2\) is rolled (an angle \(2\theta\)) along a straight line. 2ff7e9595c