Question

A child travelling at 4.0m/s crosses a line 4.0m from starting point. she continues with the steady acceleration of 0.4m/s^2 until she crosses a mark 40m from the starting point. calculate how long it takes to go from 4.0m mark to the 40m mark

Answer 1

To calculate the time it takes for the child to go from the 4.0 m mark to the 40.0 m mark under constant acceleration, we can use the kinematic equation:

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

Where:
- \( s \) is the distance traveled (from 4.0 m to 40.0 m, which is \( 40.0 m - 4.0 m = 36.0 m \))
- \( u \) is the initial velocity (which is 4.0 m/s)
- \( a \) is the acceleration (0.4 m/s²)
- \( t \) is the time in seconds

Plugging in the values into the equation:

\[
36.0 = 4.0 t + \frac{1}{2} (0.4) t^2
\]

Simplifying,

\[
36.0 = 4.0 t + 0.2 t^2
\]

Rearranging the equation gives us:

\[
0.2 t^2 + 4.0 t - 36.0 = 0
\]

To simplify calculations, we can multiply the entire equation by 5 to eliminate the decimal:

\[
t^2 + 20 t - 180 = 0
\]

Now, we can apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 20 \), and \( c = -180 \):

First, calculate the discriminant:

\[
b^2 - 4ac = 20^2 - 4 \cdot 1 \cdot (-180) = 400 + 720 = 1120
\]

Now apply the quadratic formula:

\[
t = \frac{-20 \pm \sqrt{1120}}{2 \cdot 1}
\]

Calculating \( \sqrt{1120} \):

\[
\sqrt{1120} \approx 33.5
\]

Thus, we have:

\[
t = \frac{-20 \pm 33.5}{2}
\]

Calculating the two possible values for \( t \):

1. \( t = \frac{-20 + 33.5}{2} = \frac{13.5}{2} \approx 6.75 \) seconds
2. \( t = \frac{-20 - 33.5}{2} = \frac{-53.5}{2} \) (this would be negative and not physically meaningful in this context)

The time it takes for the child to go from the 4.0 m mark to the 40.0 m mark is approximately:

\[
\boxed{6.75 \text{ seconds}}
\]