Distance in the first and second second

Problem

Distance in the first and second second. A car starts from rest and accelerates uniformly. Discuss in five lines how much distance it covers in the first second and how much in the second second, and why it covers more even though the acceleration is identical.

Solution

Starting from rest with constant acceleration $a$, the distance covered from the initial instant is:

$x(t) = \frac{1}{2}\,a\,t^2$

In the first second (from $t = 0$ to $t = 1$) the car covers:

$x(1) = \frac{1}{2}\,a\,(1)^2 = \frac{a}{2}$

By $t = 2\;\text{s}$ it has instead covered $x(2) = \tfrac{1}{2}a(2)^2 = 2a$. So in the second second (from $t = 1$ to $t = 2$) it covers only the difference:

$\Delta x = x(2) - x(1) = 2a - \frac{a}{2} = \frac{3a}{2}$

In the second interval the car covers $\tfrac{3a}{2}$, that is three times the distance of the first. The acceleration is identical, but in the second second the car already starts with a non-zero velocity (the one built up in the first second): the average velocity of the interval is larger, so the distance covered increases.

$\ev{\text{first second: } \tfrac{a}{2}; \quad \text{second second: } \tfrac{3a}{2} \; (\times 3)}$

Collegamenti

Argomenti: Kinematics Concetti: Uniformly accelerated motion