A kayaker paddles 210 m along a winding shoreline and ends 150 m east of her put-in, 50 s after starting. In v = Δx / Δt, which of the trip's numbers plays the part of Δx?
The subway train went in at 4 m/s, came out at 10 m/s, and took 6 s. You've already used Δx = ½(v₀ + v)t on stretches like that. But why does averaging the two velocities and multiplying by the time give the displacement?
The train's velocity climbs steadily: 4 m/s at the start, 7 m/s at the halfway moment, 10 m/s at the end.
The figure shows that stretch as a velocity-time graph: the train's steady climb from 4 m/s to 10 m/s draws a straight ramp.
A steady climb spends as much time below its halfway value as above it — so the train's average velocity for the whole stretch is that halfway value: ½(v₀ + v).
At constant acceleration, why does the bracket ½(v₀ + v) appear in Δx = ½(v₀ + v)t?
You've already seen that average velocity is displacement divided by elapsed time. Multiply the average velocity by the time, and you have the displacement: Δx = ½(v₀ + v)t.
On a velocity-time graph, what does the area between the line and the time axis over an interval give?
The graph tells the same story, because the area between a velocity-time graph and the time axis is the displacement, with area below the axis counting as negative. It is the same area method you've already used.
The shaded region between the train's ramp and the time axis is a trapezoid: average height 7 m/s, width 6 s, area 7 × 6 = 42 m. So ½(v₀ + v)t is exactly the trapezoid's area: the equation and the graph are two readings of one calculation.
An object speeds up steadily on a velocity-time graph, drawing a straight ramp from v₀ up to v. The displacement Δx = ½(v₀ + v)t equals which feature of that graph?
This averaging works for any motion with constant acceleration, whatever the two velocities. If the acceleration is not constant, the velocity does not climb steadily, ½(v₀ + v) is no longer the true average velocity, and the equation fails. That is the same constant-acceleration condition every kinematic equation carries.
A metro train speeds up steadily from 6 m/s to 14 m/s during 3 s. Use the average-velocity reasoning to find its displacement.
Write down the values in the question:
v₀ = 6 m/s
v = 14 m/s
t = 3 s
The velocity changes steadily, so the average velocity is the halfway value:
average velocity = ½(6 + 14) = 10 m/s
Average velocity multiplied by the time gives the displacement:
Δx = 10 × 3
Δx = 30 m, the same number the trapezoid area under the train's velocity-time ramp gives.
A van's velocity climbs steadily from 6 m/s to 12 m/s during 4 s. Using the average-velocity reasoning, which product gives its displacement?
Why does Δx = ½(v₀ + v)t require the acceleration to be constant?
An object slows steadily from 10 m/s to 4 m/s over 6 s. Which calculation gives its displacement?
For one steady slowdown from 14 m/s to 6 m/s lasting 4 s, Ana computes ½ × (14 + 6) × 4. Ben finds the area between the motion's velocity-time line and the time axis. How do the two results compare?