The diagram shows a cyclist on a bicycle and their combined mass is 120 kg - Leaving Cert Physics - Question a - 2010

To find the maximum velocity after 15 seconds, we can use the equation of motion: $v = u + at$ Where:

• $u$ is the initial velocity (0 m/s, since the cyclist starts from rest)
• $a$ is the acceleration (0.5 m/s²)
• $t$ is the time (15 s)

Substituting the values: $v = 0 + (0.5) \times (15) = 7.5 \, \text{m/s}$

Therefore, the maximum velocity of the cyclist after 15 seconds is $7.5 \, \text{m/s}$.

To calculate the distance travelled, we can use the equation: $s = ut + \frac{1}{2}at^2$ Where:

• $u$ is the initial velocity (0 m/s)
• $a$ is the acceleration (0.5 m/s²)
• $t$ is the time (15 s)

Substituting the values: $s = (0) \times (15) + \frac{1}{2} (0.5) (15^2) = \frac{1}{2} (0.5) (225) = 56.25 \, \text{m}$

Thus, the distance travelled by the cyclist during the first 15 seconds is $56.25 \, \text{m}$.

The bicycle stops due to friction and air resistance acting against its motion. Once the cyclist stops pedalling, no further net force is applied, and the opposing forces gradually bring the bicycle to a halt.

To calculate the time taken to travel the final 80 m, we assume the cyclist coasts at maximum velocity before coming to a stop. Using the formula: $s = vt$ Where:

• $s$ = 80 m
• $v$ = 7.5 m/s

Rearranging for $t$ gives: $t = \frac{s}{v} = \frac{80}{7.5} \approx 10.67 \, \text{s}$

Therefore, the time taken for the cyclist to travel the final 80 m is approximately $10.67 \, \text{s}$.