Solved problem: A bicycle with constant acceleration

Development
Formulas

Statement

difficulty

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. v₀ = 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:

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

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:

$v = v_{0} + a \cdot t \Rightarrow a = \frac{v - v_{0}}{t} \Rightarrow a = \frac{2 m / s - 0 m / s}{10 s} \Rightarrow a = 0.2 m / s^{2}$

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., v₀=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:

$v = v_{0} + a \cdot t \Rightarrow a = \frac{v - v_{0}}{t} \Rightarrow a = \frac{0 m / s - 2 m / s}{6 s} \Rightarrow a = - 0.33 m / s^{2}$

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

$x = x_{0} + v_{0} \cdot t + \frac{a \cdot t^{2}}{2} \Rightarrow x = 0 m + 0 m / s \cdot 10 s + \frac{(0.2) m / s^{2} \cdot (10 s)^{2}}{2} \Rightarrow x = 10 m$

Distance traveled in the 2nd phase

$x = x_{0} + v_{0} \cdot t + \frac{a \cdot t^{2}}{2} \Rightarrow x = 0 m + 2 m / s \cdot 6 s + \frac{(- 0.33) m / s^{2} \cdot (6 s)^{2}}{2} \Rightarrow x = 12 m - 5.94 m \Rightarrow x = 6.06 m$

Therefore, the total distance traveled is:

$x_{t o t a l} = 10 m + 6.06 m = 16.06 m$

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas

Related sections