Question

a bicyclist steadily speeds up from rest to 6.00m/s in 6.10s. how far did she travel during this time?

Explanation:

Step1: Find the acceleration

The initial velocity $u = 0m/s$, the final velocity $v=6.00m/s$ and the time $t = 6.10s$. We use the formula $v=u + at$. Rearranging for $a$ gives $a=\frac{v - u}{t}$. Substituting the values: $a=\frac{6.00 - 0}{6.10}=\frac{6.00}{6.10}\approx0.984m/s^{2}$.

Step2: Find the distance traveled

We use the second - order kinematic equation $s=ut+\frac{1}{2}at^{2}$. Since $u = 0m/s$, the equation simplifies to $s=\frac{1}{2}at^{2}$. Substituting $a=\frac{6.00}{6.10}m/s^{2}$ and $t = 6.10s$ into the equation: $s=\frac{1}{2}\times\frac{6.00}{6.10}\times(6.10)^{2}$. First, $(6.10)^{2}=37.21$, then $\frac{1}{2}\times\frac{6.00}{6.10}\times37.21 = 18.3m$.

Answer:

18.3 m