← Back to course overview

Calculating Distance and Displacement from Position and Velocity vs Time Graphs

Unit: 1D Kinematics

Lesson Preview

Consider the following diagram depicting the motion of a ball bouncing between two walls:

Loading visualization…

What is the total distance travelled by the ball?

Solution

From the position graph, we can track the object's position at each transition point:

At t=0 st = 0\text{ s}: x0=5 mx_0 = 5\text{ m}

At t=2 st = 2\text{ s}: x1=x0+v1Δt1=5+32=11 mx_1 = x_0 + v_1 \cdot \Delta t_1 = 5 + 3 \cdot 2 = 11\text{ m}

At t=5 st = 5\text{ s}: x2=x1+v2Δt2=11+(4)3=1 mx_2 = x_1 + v_2 \cdot \Delta t_2 = 11 + (-4) \cdot 3 = -1\text{ m}

At t=7 st = 7\text{ s}: x3=x2+v3Δt3=1+22=3 mx_3 = x_2 + v_3 \cdot \Delta t_3 = -1 + 2 \cdot 2 = 3\text{ m}

Displacement: Displacement is the change in position from start to finish:

Δx=xfinalxinitial=35=2 m \Delta x = x_{\text{final}} - x_{\text{initial}} = 3 - 5 = -2 \text{ m}

...

... continued in the full lesson.

Ready to Start Learning?

Sign up now to access the full Calculating Distance and Displacement from Position and Velocity vs Time Graphs lesson and our entire curriculum!