Velocity Graphing Calculator

Slopes, tangents and areas on position-time and velocity-time graphs, solved step by step
A position-time graph passes through (2.0 s, 8.0 m) and (6.0 s, 24.0 m). Find the velocity.
Find the instantaneous velocity at t = 3 s for x(t) = 3t^2 - 4t + 2
A velocity-time graph rises from 0 to 12 m/s in 4 s, holds for 6 s, then drops to 0 in 2 s. Find the displacement.
Find the acceleration from a velocity-time graph through (1 s, 4 m/s) and (5 s, 16 m/s)

Velocity From a Position-Time Graph

On a position-time graph the vertical axis is position xx in metres and the horizontal axis is time tt in seconds, so velocity is the slope, in m/s.

Average velocity between two points is the slope of the straight line joining them (the secant):

vavg=x2тИТx1t2тИТt1v_{\text{avg}} = \frac{x_2 - x_1}{t_2 - t_1}

Instantaneous velocity at one instant is the slope of the tangent there, which is the derivative:

v(t)=dxdtv(t) = \frac{dx}{dt}

How to read the shape:

  • Horizontal line тАФ at rest, v=0v = 0
  • Straight sloped line тАФ constant velocity; steeper means faster
  • Curve bending upwards тАФ speeding up in the positive direction
  • Negative slope тАФ moving in the negative direction

The assumption people forget: the secant slope and the tangent slope agree only when the graph is straight. On a curve, an average velocity over an interval is not the velocity at any particular instant inside it тАФ except at one point, by the mean value theorem.

Velocity-Time Graphs: Slope and Area

Swap the vertical axis to velocity and both features change meaning:

  • Slope is acceleration, a=╬Фv/╬Фta = \Delta v / \Delta t, in m/s┬▓
  • Area under the curve is displacement, in metres

╬Фx=тИлt1t2v(t)тАЙdt\Delta x = \int_{t_1}^{t_2} v(t)\,dt

In practice you rarely integrate. Break the shape into rectangles and triangles: a rectangle contributes vтЛЕ╬Фtv \cdot \Delta t, a triangle contributes 12╬ФvтЛЕ╬Фt\tfrac12 \Delta v \cdot \Delta t.

Area below the time axis counts as negative тАФ the object is moving backwards, so it subtracts from the displacement.

When it applies: the area rule is exact for any velocity curve, straight or not; it is nothing more than integration.

The assumption people forget: the signed area gives displacement, not distance travelled. For distance, add the absolute values of the areas above and below the axis separately.

Common Mistakes to Avoid

  • Reading the height of a position-time graph as velocity тАФ the height is position. Velocity is the slope.
  • Mixing up the two graph types тАФ a horizontal line means "at rest" on a position-time graph but "constant velocity" on a velocity-time graph.
  • Using a tangent slope as an average тАФ the tangent gives one instant only.
  • Forgetting the sign of the area тАФ a trip out and back has a positive area then a negative one, and they cancel to zero displacement.
  • Ignoring the axis scale тАФ a squashed vertical axis makes a gentle slope look dramatic. Always compute from the numbers, not the visual steepness.
  • Assuming a curved position-time graph means changing acceleration тАФ a parabola means constant acceleration.

Examples

Step 1: The graph is straight, so the velocity is constant and equals the slope
Step 2: v=x2тИТx1t2тИТt1=24.0┬аmтИТ8.0┬аm6.0┬аsтИТ2.0┬аsv = \dfrac{x_2 - x_1}{t_2 - t_1} = \dfrac{24.0\ \text{m} - 8.0\ \text{m}}{6.0\ \text{s} - 2.0\ \text{s}}
Step 3: v=(16.0┬аm)├╖(4.0┬аs)=4.0┬аm/sv = (16.0\ \text{m}) \div (4.0\ \text{s}) = 4.0\ \text{m/s}
Answer: v=4.0v = 4.0 m/s

Step 1: v(t)=dx/dt=6tтИТ4v(t) = dx/dt = 6t - 4, in m/s
Step 2: v(3.0)=6(3.0)тИТ4=18тИТ4=14┬аm/sv(3.0) = 6(3.0) - 4 = 18 - 4 = 14\ \text{m/s}
Step 3: x(1.0)=3(1)тИТ4(1)+2=1.0┬аmx(1.0) = 3(1) - 4(1) + 2 = 1.0\ \text{m}; x(3.0)=3(9)тИТ4(3)+2=17.0┬аmx(3.0) = 3(9) - 4(3) + 2 = 17.0\ \text{m}
Step 4: vavg=(17.0┬аmтИТ1.0┬аm)├╖(3.0┬аsтИТ1.0┬аs)=16.0├╖2.0=8.0┬аm/sv_{\text{avg}} = (17.0\ \text{m} - 1.0\ \text{m}) \div (3.0\ \text{s} - 1.0\ \text{s}) = 16.0 \div 2.0 = 8.0\ \text{m/s}
Answer: v(3.0┬аs)=14v(3.0\ \text{s}) = 14 m/s; vavg=8.0v_{\text{avg}} = 8.0 m/s over the interval

Step 1: Phase 1 (triangle): 12(4.0┬аs)(12┬аm/s)=24┬аm\tfrac12 (4.0\ \text{s})(12\ \text{m/s}) = 24\ \text{m}; a=12├╖4.0=3.0┬аm/s2a = 12 \div 4.0 = 3.0\ \text{m/s}^2
Step 2: Phase 2 (rectangle): (12┬аm/s)(6.0┬аs)=72┬аm(12\ \text{m/s})(6.0\ \text{s}) = 72\ \text{m}; a=0a = 0
Step 3: Phase 3 (triangle): 12(2.0┬аs)(12┬аm/s)=12┬аm\tfrac12 (2.0\ \text{s})(12\ \text{m/s}) = 12\ \text{m}; a=тИТ12├╖2.0=тИТ6.0┬аm/s2a = -12 \div 2.0 = -6.0\ \text{m/s}^2
Step 4: Total displacement =24┬аm+72┬аm+12┬аm=108┬аm= 24\ \text{m} + 72\ \text{m} + 12\ \text{m} = 108\ \text{m}
Answer: ╬Фx=108\Delta x = 108 m; accelerations 3.03.0, 00 and тИТ6.0-6.0 m/s┬▓

Frequently Asked Questions

Take the slope. For average velocity between two times, use the rise over run of the straight line joining those points: v = (xтВВ тИТ xтВБ)/(tтВВ тИТ tтВБ). For instantaneous velocity, draw the tangent at that instant and take its slope, which is the derivative dx/dt.

Displacement, in metres. Split the shape into rectangles and triangles and add the areas, counting anything below the time axis as negative. Adding the absolute values instead gives the total distance travelled.

Acceleration, in m/s┬▓. A straight sloped line means constant acceleration, a horizontal line means constant velocity with zero acceleration, and a curve means the acceleration itself is changing.

Compare the signs of velocity and acceleration on the velocity-time graph. Same sign means speeding up, opposite signs means slowing down. A line heading toward the time axis is slowing down whether it sits above or below the axis.

Related Solvers

Try AI-Math for Free

Get step-by-step solutions to any math problem. Upload a photo or type your question.

Start Solving