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 3.00° to the horizontal. The child starts partway down the hill, 7.00 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 3.70 m/s^2.

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

the child's acceleration is ... g * sin(3º)

1/2 * 3.70 * t^2 = 1/2 * [9.81 sin(3º)] * t^2 + 7.00

solve for t , substitute back to find the child's travel distance

To calculate how far the child has gone before her parent catches her, we need to find the time it takes for the parent to reach the child. Then we can use the time to calculate the distance.

First, let's find the time it takes for the parent to reach the child. We can use the kinematic equation:

v = u + at

where:
v = final velocity (which is the velocity of the parent when he catches the child)
u = initial velocity (0 m/s because the parent starts from rest)
a = acceleration of the parent (3.70 m/s^2)
t = time

Since we're looking for the time it takes for the parent to reach the child, we rearrange the equation:

t = (v - u) / a

The final velocity can be found using another kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity
u = initial velocity (0 m/s because the parent starts from rest)
a = acceleration of the parent (3.70 m/s^2)
s = distance covered by the parent (which is the same as the distance the child has gone)

Rearranging the equation to solve for v:

v = sqrt(u^2 + 2as)

Now we can substitute the values into the equations:

v = sqrt(0^2 + 2 * 3.70 * 7.00) = 7.74 m/s (rounded to two decimal places)

Now we can find the time:

t = (7.74 - 0) / 3.70 = 2.09 seconds (rounded to two decimal places)

Finally, we can calculate the distance the child has gone before her parent catches her:

s = u * t + 0.5 * a * t^2

where:
u = initial velocity (0 m/s because the child starts from rest)
a = acceleration of the child due to the incline (we'll find it later)
t = time (2.09 seconds)

To find the acceleration of the child, we can use trigonometry. The acceleration due to the incline can be calculated using the formula:

a_incline = g * sin(θ)

where:
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of the incline (3.00°)

Plugging in the values:

a_incline = 9.8 * sin(3.00°) = 0.51 m/s^2 (rounded to two decimal places)

Now we can calculate the distance the child has gone:

s_child = 0 * 2.09 + 0.5 * 0.51 * (2.09)^2 = 1.27 meters (rounded to two decimal places)

Therefore, the child has gone approximately 1.27 meters before her parent catches her.

35