12. Answer any two of the following parts (a), (b), (c), (d) - Leaving Cert Physics - Question 12 - 2010

To find the maximum velocity, we use the equation:

$v = u + at$

Where:

• $u = 0 \, m/s$ (initial velocity)
• $a = 0.5 \, m/s^2$ (acceleration)
• $t = 15 \, s$ (time)

Calculating:

$v = 0 + (0.5)(15) = 7.5 \, m/s$

The maximum velocity of the cyclist after 15 seconds is $7.5 \, m/s$.

To find the distance, we use the equation:

$s = ut + \frac{1}{2}at^2$

Where:

• $u = 0 \, m/s$ (initial velocity)
• $a = 0.5 \, m/s^2$ (acceleration)
• $t = 15 \, s$

Calculating the distance:

$s = (0)(15) + \frac{1}{2}(0.5)(15^2) = 0 + 56.25 = 56.25 \, m$

The distance travelled by the cyclist during the first 15 seconds is $56.25 \, m$.

The bicycle stops due to friction and air resistance acting against the motion of the cyclist. These forces eventually overcome the cyclist's momentum, leading to a gradual decrease in speed until the bicycle comes to a halt.

To find the time taken for the cyclist to travel the final 80 m, we can use the equation:

$s = ut + \frac{1}{2}at^2$

Here:

• $u = 7.5 \, m/s$ (maximum velocity when pedalling ends)
• $s = 80 \, m$
• $a = 0 \, m/s^2$ (deceleration is negligible while coasting)

This simplifies to: