Estimates from the x(t) graph

Problem

The graph in the figure shows the position $x(t)$ of a body. Estimate:

• (a) the average velocity between $t = 0$ and $t = 4\;\text{s}$;
• (b) the instant at which the instantaneous velocity is zero;
• (c) the instant at which the velocity is maximum in magnitude.

Solution

How to read an $x(t)$ graph. The instantaneous velocity is the slope (the inclination of the tangent) to the curve; the average velocity between two instants is the slope of the chord joining the two corresponding points.

(a) Average velocity between $t = 0$ and $t = 4\;\text{s}$. We read the two positions off the graph: $x(0) \approx 0{.}5\;\text{m}$ and, at the maximum of the curve, $x(4) \approx 2{.}1\;\text{m}$. Hence

$\bar v = \frac{x(4) - x(0)}{4 - 0} \approx \frac{2{.}1 - 0{.}5}{4} \approx 0{.}4\;\text{m/s}$

(b) Instant of zero velocity. The instantaneous velocity is zero where the tangent is horizontal, i.e. where the curve reaches its maximum. From the graph this occurs at

$\ev{t \approx 4\;\mathrm{s}}$

Before this instant the body moves forward ($x$ increases), afterwards it goes back ($x$ decreases): the turning point is the maximum.

(c) Instant of maximum velocity in magnitude. The velocity is maximum in magnitude where the curve is steepest. In the rising phase the slope is maximum around $t \approx 2\;\text{s}$ ($v \approx 0{.}6\;\text{m/s}$); after the maximum, in the return phase, the descent grows steadily steeper towards the end of the interval. The largest slope in magnitude over the whole graph is therefore near

$\ev{t \approx 5\;\mathrm{s}, \quad |v| \approx 0{.}7\;\mathrm{m/s}}$

Method note. All these values are estimates read off the graph: the aim of the exercise is to learn to derive velocities (slopes) from a position-time curve, not to obtain exact figures.

Collegamenti

Argomenti: Kinematics Concetti: Average velocity · Acceleration Competenze: Graph reading