On a velocity-versus-time graph, which quantity equals the object's displacement?

On a velocity-versus-time graph, which quantity equals the object's displacement?

• A. The maximum velocity
• B. The intercept with the t-axis
• C. The slope of the graph
• D. The area between the graph and the t-axis

Answer: (D)

Displacement is the net change in position, which comes from velocity being the rate at which position changes. Over a small interval, the position change is Δx ≈ v Δt. If you sum up all those tiny changes over a time interval, you’re effectively integrating velocity over time. Graphically, that sum corresponds to the area under the velocity-versus-time curve between the start and end times. If velocity stays positive, the area adds to displacement; if velocity goes negative, that area subtracts, giving the net displacement. The maximum velocity tells you how fast you’re going at a moment, the intercept with the t-axis indicates when velocity is zero, and the slope of the graph is acceleration. The area under the v–t graph is the quantity that matches the object's displacement.

Displacement is the net change in position, which comes from velocity being the rate at which position changes. Over a small interval, the position change is Δx ≈ v Δt. If you sum up all those tiny changes over a time interval, you’re effectively integrating velocity over time. Graphically, that sum corresponds to the area under the velocity-versus-time curve between the start and end times. If velocity stays positive, the area adds to displacement; if velocity goes negative, that area subtracts, giving the net displacement. The maximum velocity tells you how fast you’re going at a moment, the intercept with the t-axis indicates when velocity is zero, and the slope of the graph is acceleration. The area under the v–t graph is the quantity that matches the object's displacement.