Problem — From distance to time

Problem

Inverse: given the distance, find the time. A car starts from rest with acceleration $2{,}5\;\text{m/s}^2$. After how much time has it covered $100\;\text{m}$? Also calculate the final speed.

Solution

Setup. The motion is uniformly accelerated starting from rest ($v_0 = 0$), so the distance covered grows with the square of the time:

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

Finding the time. We isolate $t$ by imposing $x = 100\;\text{m}$:

$t = \sqrt{\frac{2x}{a}} = \sqrt{\frac{2\cdot 100}{2{,}5}} = \sqrt{80}\;\text{s}$

$\ev{t \approx 8{,}94\;\mathrm{s}}$

Final speed. In uniformly accelerated motion the speed grows linearly with time, $v = a\,t$:

$v = a\,t = 2{,}5 \cdot 8{,}94 \approx 22{,}4\;\text{m/s}$

$\ev{v_\text{fin} \approx 22{,}4\;\mathrm{m/s}}$

As a check we can use the relation without time $v^2 = 2ax = 2\cdot 2{,}5\cdot 100 = 500$, from which $v = \sqrt{500} \approx 22{,}4\;\text{m/s}$: consistent.

Links

Topics: Cinematica Concepts: Moto uniformemente accelerato