The graphs in Figure 3 below represent the motion of the jet-powered car as it accelerates toward its top speed, but only during the time when its acceleration is constant. Time starts at zero for this motion (as if measured with a stopwatch), and the displacement and velocity are initially 200 m and 15 m/s, respectively.

The graph of displacement versus time in Figure 3(a) is a curve rather than a straight line. The slope of the curve becomes steeper as time progresses, showing that the velocity is increasing over time. The slope at any point on a displacement-versus-time graph is the instantaneous velocity at that point. It is found by drawing a straight line tangent to the curve at the point of interest and taking the slope of this straight line. Tangent lines are shown for two points in Figure 3(a). If this is done at every point on the curve and the values are plotted against time, then the graph of velocity versus time shown in Figure 3(b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in Figure 3(c).

Calculate the velocity of the jet car at a time of 25 s by finding the slope of the xx size 12{x} {} vs. tt size 12{t} {} graph in the graph below.

*Strategy*

The slope of a curve at a point is equal to the slope of a straight line tangent to the curve at that point. This principle is illustrated in Figure 5, where Q is the point at t=25 st=25 s size 12{t="25"`s} {}.

*Solution*

1. Find the tangent line to the curve at t=25 st=25 s size 12{t="25"`s} {}.

2. Determine the endpoints of the tangent. These correspond to a position of 1300 m at time 19 s and a position of 3120 m at time 32 s.

3. Plug these endpoints into the equation to solve for the slope, *vv size 12{v} {}*.

slope
=
v
Q
=
Δx
Q
Δt
Q
=
3120 m
−
1300 m
32 s
−
19 s
slope
=
v
Q
=
Δx
Q
Δt
Q
=
3120 m
−
1300 m
32 s
−
19 s
size 12{"slope"=v rSub { size 8{Q} } = { {Δx rSub { size 8{Q} } } over {Δt rSub { size 8{Q} } } } = { { left ("3120"`m - "1300"`m right )} over { left ("32"`s - "19"`s right )} } } {}

(8)Thus,

v
Q
=
1820 m
13 s
=
140 m/s.
v
Q
=
1820 m
13 s
=
140 m/s.

(9)
*Discussion*

This is the value given in this figure’s table for vv size 12{v} {} at t=25 st=25 s. The value of 140 m/s for vQvQ is plotted in Figure 5. The entire graph of vv vs. tt can be obtained in this fashion.

Carrying this one step further, we note that the slope of a velocity versus time graph is acceleration. Slope is rise divided by run; on a vv size 12{v} {} vs. tt graph, rise = change in velocity ΔvΔv size 12{Dv} {} and run = change in time ΔtΔt size 12{Dt} {}.

The slope of a graph of velocity vv size 12{v} {} vs. time tt size 12{t} {} is acceleration aa size 12{a} {}*.*

slope
=
Δv
Δt
=
a
slope
=
Δv
Δt
=
a

(10) Since the velocity versus time graph in Figure 3(b) is a straight line, its slope is the same everywhere, implying that acceleration is constant. Acceleration versus time is graphed in Figure 3(c).

Additional general information can be obtained from Figure 5 and the expression for a straight line, y=mx+by=mx+b size 12{y= ital "mx"+b} {}.

In this case, the vertical axis yy size 12{y} {} is VV size 12{V} {}, the intercept bb size 12{b} {} is v0v0 size 12{v rSub { size 8{0} } } {}, the slope mm size 12{m} {} is aa size 12{a} {}, and the horizontal axis xx size 12{x} {} is tt size 12{t} {}. Substituting these symbols yields

v=v0+at.v=v0+at. size 12{v=v rSub { size 8{0} } + ital "at"} {}

(11)A general relationship for velocity, acceleration, and time has again been obtained from a graph. Notice that this equation was also derived algebraically from other motion equations in Motion Equations for Constant Acceleration in One Dimension.

It is not accidental that the same equations are obtained by graphical analysis as by algebraic techniques. In fact, an important way to *discover* physical relationships is to measure various physical quantities and then make graphs of one quantity against another to see if they are correlated in any way. Correlations imply physical relationships and might be shown by smooth graphs such as those above. From such graphs, mathematical relationships can sometimes be postulated. Further experiments are then performed to determine the validity of the hypothesized relationships.

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