Question

A student is running at her top speed of 5.2 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 41.5 m from the bus, it starts to pull away, moving with a constant acceleration of 0.171 m/s2 .

What is the minimum speed the student must have to just catch up with the bus?

Answer 1

((The way I will do it, is through calculus.))

Function for the bus displacement: (41.5
is the initial distance of bus from student)

\(\color{#0000ff }{ \displaystyle {\rm x}_1(t)=(0.171/2)t^2+41.5}\)

Student's function for displacement:
(We don't know the velocity the student goes at
when bus starts to pull away from the bus-stop)

\(\color{#0000ff }{ \displaystyle {\rm x}_2(t)={\bf v}{\tiny~}t}\)

When can the functions intersect?

\(\color{#0000ff }{ \displaystyle {\rm x}_1(t)={\rm x}_2(t)}\)
\(\color{#0000ff }{ \displaystyle 0.0855{\tiny~}t^2+41.5={\bf v}{\tiny~}t}\)
\(\color{#0000ff }{ \displaystyle {\bf v}=0.0855t+(41.5/t)}\)

Let's minimize this function! Recall the constraints
given in the problem, and by definition of "time".
\(\color{#0000ff }{ \displaystyle 0\le {\bf v}\le 5.2~~{m/s}}\) and \(\color{#0000ff }{ \displaystyle t>0~~{s}}\)

I will minimize the function:
\(\color{#0000ff }{ \displaystyle {\bf v}'=0.0855-(41.5/t^2)}\)

the only critical number is:
\(\color{#0000ff }{ \displaystyle t=22~m/s}\)
(this is the time when min velocity is reached)

\(\color{#0000ff }{ \displaystyle {\bf v}_{\rm min}={\bf v}(22)=3.77~m/s}\)

So, I get that \(\color{#ff0000}{ \displaystyle {\bf v}_{\rm min}=3.77~m/s}\) as the answer.

Answer 2

how did you get the function for the bus displacement? I'm a bit confused about that thanks