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Graphs of motion when a size 12{a} {} Is constant but a 0 size 12{a<>0} {}

The graphs in [link] 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.

Three line graphs. First is a line graph of displacement over time. Line has a positive slope that increases with time. Second line graph is of velocity over time. Line is straight with a positive slope. Third line graph is of acceleration over time. Line is straight and horizontal, indicating constant acceleration.
Graphs of motion of a jet-powered car during the time span when its acceleration is constant. (a) The slope of an x size 12{x} {} vs. t size 12{t} {} graph is velocity. This is shown at two points, and the instantaneous velocities obtained are plotted in the next graph. Instantaneous velocity at any point is the slope of the tangent at that point. (b) The slope of the v size 12{v} {} vs. t size 12{t} {} graph is constant for this part of the motion, indicating constant acceleration. (c) Acceleration has the constant value of 5 . 0 m/s 2 size 12{5 "." "0 m/s" rSup { size 8{2} } } {} over the time interval plotted.
A U.S. Air Force jet car speeds down a track. (credit: Matt Trostle, Flickr)

The graph of displacement versus time in [link] (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 [link] (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 [link] (b) is obtained. Furthermore, the slope of the graph of velocity versus time is acceleration, which is shown in [link] (c).

Determining instantaneous velocity from the slope at a point: jet car

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

A graph of displacement versus time for a jet car. The x axis for time runs from zero to thirty five seconds. The y axis for displacement runs from zero to three thousand meters. The curve depicting displacement is concave up. The slope of the curve increases over time. Slope equals velocity v. There are two points on the curve, labeled, P and Q. P is located at time equals ten seconds. Q is located and time equals twenty-five seconds. A line tangent to P at ten seconds is drawn and has a slope delta x sub P over delta t sub p. A line tangent to Q at twenty five seconds is drawn and has a slope equal to delta x sub q over delta t sub q. Select coordinates are given in a table and consist of the following: time zero seconds displacement two hundred meters; time five seconds displacement three hundred thirty eight meters; time ten seconds displacement six hundred meters; time fifteen seconds displacement nine hundred eighty eight meters. Time twenty seconds displacement one thousand five hundred meters; time twenty five seconds displacement two thousand one hundred thirty eight meters; time thirty seconds displacement two thousand nine hundred meters.
The slope of an x size 12{x} {} vs. t size 12{t} {} graph is velocity. This is shown at two points. Instantaneous velocity at any point is the slope of the tangent at that point.


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 [link] , where Q is the point at t = 25 s size 12{t="25"`s} {} .


1. Find the tangent line to the curve at t = 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, v size 12{v} {} .

slope = v Q = Δ x Q Δ t 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 )} } } {}


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


This is the value given in this figure's table for v size 12{v} {} at t = 25 s . The value of 140 m/s for v Q is plotted in [link] . The entire graph of v vs. t can be obtained in this fashion.

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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 v size 12{v} {} vs. t graph, rise = change in velocity Δ v size 12{Dv} {} and run = change in time Δ t size 12{Dt} {} .

The slope of v Vs. t

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

slope = Δ v Δ t = a

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

Questions & Answers

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Practice Key Terms 4

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