## Statement

A cyclist starts his morning ride and after 10 seconds his velocity is 7.2 km/h. At that moment, he sees a dog approaching and slows down for 6 seconds until the bicycle stops. Calculate:

a) The acceleration until he begins to slow down.

b) The braking acceleration of the bike.

c) The total distance traveled.

## Solution

The movement can be broken down in 2 phases. A first phase in which the acceleration is positive (a > 0) and the second where the acceleration is negative, since he is braking (a < 0)

**Question a)**

**Data**

*Initial velocity*. v_{0} = 0 m/s

*Velocity at 10 s*. v = 7.2 km/h.

Transforming the velocity to S.I. units, we have that the velocity at 10 s is:

$$v=7.2\overline{)km}/\overline{)h}*\frac{1000m}{1\overline{)km}}*\frac{1\overline{)h}}{3600s}=2m/s$$

**Resolution**

We are asked for the acceleration in the first phase of the motion. Since we know the initial velocity (0 m/s), the final velocity (2 m/s) and the time between them (10 s), we can use the velocity equation and calculate the acceleration directly:

$$v={v}_{0}+a\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}a=\frac{v-{v}_{0}}{t}\Rightarrow \phantom{\rule{0ex}{0ex}}a=\frac{2m/s-0m/s}{10s}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)a=0.2m/{s}^{2}}$$

**Question b)**

In this question, we are asked for the acceleration in the second phase.

**Data**

The *initial velocity* would be the final velocity of the first phase, i.e., v_{0}=2m/s.

*Velocity at 6 s*. Since he stops, at that moment, the velocity will be 0: v=0 m/s.

**Resolution**

Applying the same equation as in question a, we obtain:

$$v={v}_{0}+a\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}a=\frac{v-{v}_{0}}{t}\Rightarrow \phantom{\rule{0ex}{0ex}}a=\frac{0m/s-2m/s}{6s}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)a=-0.33m/{s}^{2}}$$

**Question c)**

The distance traveled by the cyclist will be the distance traveled in the first phase plus the distance traveled in the second.

Distance traveled in the 1st phase

$$x={x}_{0}+{v}_{0}\xb7t+\frac{a\xb7{t}^{2}}{2}\Rightarrow \phantom{\rule{0ex}{0ex}}x=0m+0m/s\xb710s+\frac{(0.2)m/{s}^{2}\xb7(10s{)}^{2}}{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)x=10m}$$

Distance traveled in the 2nd phase

$$x={x}_{0}+{v}_{0}\xb7t+\frac{a\xb7{t}^{2}}{2}\Rightarrow \phantom{\rule{0ex}{0ex}}x=0m+2m/\overline{)s}\xb76\overline{)s}+\frac{(-0.33)m/{s}^{2}\xb7(6s{)}^{2}}{2}\Rightarrow \phantom{\rule{0ex}{0ex}}x=12m-5.94m\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)x=6.06m}$$

Therefore, the total distance traveled is:

$${x}_{total}=10m+6.06m=\overline{)16.06m}$$