Which graph x(t)

Problem

Which $x(t)$? A car decelerates uniformly until it stops, starting from a positive initial velocity. Which of the graphs $x(t)$ below represents its motion? (Epstein 2002)

Solution

The key is the slope. In a position-time graph, the slope of the curve is the instantaneous velocity. Let’s translate the text into conditions on the slope:

• “positive initial velocity” $\Rightarrow$ at the initial instant the curve rises with a net, positive slope;
• “decelerates uniformly” $\Rightarrow$ the slope decreases steadily (the curve is concave down);
• “until it stops” $\Rightarrow$ at the end the slope goes to zero, i.e. the curve becomes horizontal.

Analysis of the three graphs.

• A — parabola with increasing slope (concave up): represents a motion that is accelerating, not slowing down. Discarded.
• B — parabola that starts steep and flattens out until it becomes horizontal: the slope (the velocity) starts large and decreases to zero. This is exactly a uniform deceleration to a stop. Correct.
• C — straight line with constant slope: constant velocity, no deceleration. Discarded.

Algebraic check. With $v_0 > 0$ and constant $a < 0$ the equation of motion is $x(t) = x_0 + v_0 t + \tfrac{1}{2} a t^2$: a downward-concave parabola whose tangent becomes horizontal at the stopping instant $t = -v_0/a$. Exactly the behaviour of B.

$\ev{\text{The correct graph is } B}$

Collegamenti

Argomenti: Kinematics Concetti: Uniformly accelerated motion Competenze: Graph reading