2 A position-time graph for a particle moving along the $x$ axis is
3 shown in Figure~P2.5. \Part{a} Find the average velocity in the time
4 interval $t=1.50\U{s}$ to $t=4.00\U{s}$. \Part{b} Determine the
5 instantaneous velocity at $t=2.00\U{s}$ by measuring the slope of the
6 tangent to the graph. \Part{c} At what value of $t$ is the velocity
12 size(4cm, IgnoreAspect);
14 /* Parameter computation from the back-of-the-book answers:
15 * 2*a*2 + b = -3.8 (1) (answer to part b)
16 * 2*a*4 + b = 0 (2) (answer to part c)
17 * 2*a*2 = 3.8 (3) ((2)-(1))
18 * a = 3.8/4 = 0.95 ~= 1.0 (4) (solve (3), rounding to likely value)
19 * b = -8*a (5) (rearrange (2))
20 * b = -8*1 = -8 (6) (plug (4) into (5))
21 * a*4^2 + b*4 + c = 2 (7) (location of minimum from graph)
22 * c = 2 - 16a - 4b (8) (rearrange (7))
23 * c = 2 - 16*1 + 4*8 = 18 (9) (plug (4) and (6) into (8))
29 return a*t*t + b*t + c;
33 real slope = 2*a*tanj_time + b;
34 return slope * (t - tanj_time) + parab(tanj_time);
37 draw(graph(tanj, 0, 3.5), green);
38 draw(graph(parab, 1, 6.5), red);
40 xaxis("$t\U{s}$", BottomTop, LeftTicks(extend=true, ptick=gray+thin()));
41 yaxis("$x\U{m}$", LeftRight, RightTicks(extend=true, ptick=gray+thin()));
48 The average velocity is the total displacement over the elapsed time, so
50 v_\text{avg} = \frac{x(4.00\U{s}) - x(1.50\U{s})}{4.00\U{s}-1.50\U{s}}
51 = \frac{2.0\U{m} - 8.0\U{m}}{2.5\U{s}}
56 Using rise-over-run to determine the tangent slope
58 v(t=2.00\U{s}) = \frac{0\U{m} - 11.5\U{m}}{3.5\U{s} - 0.5\U{s}}
63 Looking for the minimum of $x(t)$ (where the tangent curve is flat),
64 we see that $v(\ans{4.0\U{s}})=0$.