A small child is learning to ride a bike for the first time. Her dad decides it would be a good idea to start her on a small hill. The hill is inclined at 3.30° to the horizontal. The child starts partway down the hill, 9.00 m away from her dad, who stays at the top. She starts to wobble, so her dad sprints toward her from rest with an acceleration of 4.50 m/s2.

How far has the child gone before her parent catches her?

To find out how far the child has gone before her parent catches her, we first need to determine the time it takes for the parent to reach the child. We can use the kinematic equation:

\[S = ut + \frac{1}{2} a t^2\]

Where:
- \(S\) is the distance covered by an object (child or parent) in meters
- \(u\) is the initial velocity of an object in meters per second (0 m/s for the parent)
- \(a\) is the acceleration of an object in meters per second squared (4.50 m/s² for the parent)
- \(t\) is the time taken by an object in seconds

For the parent, \(u = 0\) and \(a = 4.50 \, \text{m/s}^2\). The parent starts at rest, so the initial velocity is zero. We can rearrange the equation to solve for time:

\[S = \frac{1}{2} a t^2\]
\[t^2 = \frac{2S}{a}\]
\[t = \sqrt{\frac{2S}{a}}\]

Now let's calculate the time taken by the parent to reach the child.

\[t = \sqrt{\frac{2 \times 9.00 \, \text{m}}{4.50 \, \text{m/s}^2}}\]
\[t = \sqrt{\frac{18.00}{4.50}}\]
\[t = \sqrt{4.00}\]
\[t = 2.00 \, \text{s}\]

The parent takes 2.00 seconds to reach the child.

Now, let's find the distance the child has gone in those 2.00 seconds.

The child starts partway down the hill, 9.00 m away from her dad. So the distance covered by the child is,

\[S_{child} = S_{total} - S_{parent} = 9.00 \, \text{m} - (0.5 \times a \times t^2)\]

Substituting the values,

\[S_{child} = 9.00 \, \text{m} - (0.5 \times 4.50 \, \text{m/s}^2 \times (2.00 \, \text{s})^2)\]
\[S_{child} = 9.00 \, \text{m} - (0.5 \times 4.50 \, \text{m/s}^2 \times 4.00 \, \text{s}^2)\]
\[S_{child} = 9.00 \, \text{m} - (9.00 \, \text{m})\]
\[S_{child} = 0.00 \, \text{m}\]

The child has not gone any distance in those 2.00 seconds before her dad catches her.

100m

45

2.13m