Question

two friends see each other in a store initially they are 50 m apart the first starts walking towards the second friend at a constant speed of 0.50 m s at the same instant the second friend accelerates from the rest at a rate of 1.0 m s squared tour the first friend how long before the two friends meet use quadratic formula to solve this problem

Answer 1

Let's take the origin where the first friend began. Since this friend started walking with a constant speed of 0.5 m/s and is walking towards the friend (that we are going to assume is to the right) we have that its position at any given time is:

`x=0.5t`

The position of the second friend starts at 50 m to the right and its motion is to the left, since it is walking with costant acceleration its position at any given time is:

`x=50-(1)/(2)(1)t^2`

to determine how long it takes for them to meet we equate the positions and solve for t:

`\begin{gathered} 0.5t=50-(1)/(2)(1)t^2 \\ t=100-t^2 \\ t^2+t-100=0 \\ t=\frac{-1\pm\sqrt[]{1^2-4(1)(-100)}}{2(1)} \\ t=\frac{-1\pm\sqrt[]{401}}{2} \\ \text{then} \\ t=\frac{-1+\sqrt[]{401}}{2}=9.51 \\ \text{ or} \\ t=\frac{-1-\sqrt[]{401}}{2}=-10.51 \end{gathered}`

Since time has to be positive we choose the positive solution for the equation.

Theredore, we conclude that it took 9.51 s for the friends to meet.