AP Calculus AB · Free-response skill

Reading a velocity graph

The move. From a graph of velocity made of straight pieces, find a position, a total distance and an average value from areas, and set up and evaluate an area between two curves.

How to do it

  1. Split the graph where it crosses the t-axis and where its pieces change. Find each area with geometry: triangles, rectangles and trapezoids. Mark each area as positive (above the axis) or negative (below).
  2. Position at time b is the starting position plus the signed areas up to b. Total distance adds the sizes of the areas and ignores their signs. Write the integral you are evaluating, such as ∫0bv(t) dt\int_0^b v(t)\,dt, before giving its value.
  3. Average value of the velocity on an interval is the signed area divided by the length of the interval. Give it in the velocity's units.
  4. For the area between a graph and another curve, find where they meet, check which is on top by testing a point between, and write ∫(top−bottom) dt\int (\text{top} - \text{bottom})\,dt with those limits. When one curve is made of pieces, split the integral at the corners, or use geometry on each piece.
  5. Every value here comes from geometry or a simple antiderivative, so practice this kind of question without a calculator.

Practice prompt Try it

Particle A moves along a straight track. Its velocity v(t)v(t), in meters per second, is shown in the graph for 0≤t≤100 \le t \le 10 seconds. At t=0t = 0, particle A is at position x=5x = 5 meters. A second particle, B, has velocity w(t)=2t3w(t) = \frac{2t}{3} meters per second.
Velocity v(t), in meters per second, of particle A, for 0 \le t \le 10 seconds12345678910−5−4−3−2−1123450t (seconds)v(t)(0, 0)(2, 4)(6, 4)(8, 0)(10, −4)
Velocity v(t)v(t), in meters per second, of particle A, for 0≤t≤100 \le t \le 10 seconds Three straight pieces. The graph rises in a straight line from (0, 0) to (2, 4). It stays level at 4 from t = 2 to t = 6. It then falls in a straight line from (6, 4) through (8, 0) to (10, -4). It is above the t-axis for 0 < t < 8 and below it for 8 < t < 10.

Use the graph and the information above to answer each part.

  1. Find the position of particle A at t=10t = 10. Show the integral you use.
  2. Find how many meters particle A travels in all, forward and backward, from t=0t = 0 to t=10t = 10.
  3. Find the average value of v(t)v(t) on the interval 0≤t≤80 \le t \le 8, with units.
  4. The graphs of vv and ww meet at t=0t = 0 and t=6t = 6, and v(t)≥w(t)v(t) \ge w(t) between them. Write an integral for the area of the region between the two graphs for 0≤t≤60 \le t \le 6, and find its value.

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Checklist

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Model answer, annotated

(a) The signed areas are 4 on [0,2][0, 2], 16 on [2,6][2, 6], 4 on [6,8][6, 8] and −4-4 on [8,10][8, 10], so ∫010v(t) dt=4+16+4−4=20\int_0^{10} v(t)\,dt = 4 + 16 + 4 - 4 = 20. The position at t=10t = 10 is x(10)=5+∫010v(t) dt=5+20=25x(10) = 5 + \int_0^{10} v(t)\,dt = 5 + 20 = 25 meters.

  • Point 1 Writes the integral, finds the signed areas with geometry and adds the starting position.

(b) Particle A moves forward a distance of 4+16+4=244 + 16 + 4 = 24 meters on [0,8][0, 8] and backward 4 meters on [8,10][8, 10]. The total distance is ∫010∣v(t)∣ dt=24+4=28\int_0^{10} |v(t)|\,dt = 24 + 4 = 28 meters.

  • Point 2 Adds the sizes of the areas, so the backward motion counts instead of canceling.

(c) The average value is 18−0∫08v(t) dt=248=3\frac{1}{8 - 0}\int_0^8 v(t)\,dt = \frac{24}{8} = 3 meters per second.

  • Point 3 Divides the signed area on the interval by its length and keeps the velocity units.

(d) The area is ∫06(v(t)−2t3)dt=∫06v(t) dt−∫062t3 dt=20−[t23]06=20−12=8\int_0^6 \left(v(t) - \frac{2t}{3}\right)dt = \int_0^6 v(t)\,dt - \int_0^6 \frac{2t}{3}\,dt = 20 - \left[\frac{t^2}{3}\right]_0^6 = 20 - 12 = 8.

  • Point 4 Writes top minus bottom with the meeting points as limits.
  • Point 5 Uses the graph for the first integral and an antiderivative for the second.