Question

Romeo is at `x = 0 m` at `t = 0 s` when he sees Juliet at `x = 6 m`... ?

a) Romeo begins to run towards her at `v = 5 m/s`. Juliet, in turn, begins to accelerate towards
him at `a = −2 m/s^2`. When and where will they cross?

(b) Suppose, instead, that Juliet moved away from Romeo with positive acceleration `a`. Find
`a_max`, the maximum acceleration for which Romeo can catch up with her. For this case find
the time `t` of their meeting.

I only need help with Part B, but I included Part A in case it's helpful.

Answer 1

This is what I get

Explanation:

(b) We use the kinematic expression

`s=s_0+ut+1/2at^2`

For Romeo
`s_R=0+5t`
`s_R=5t` .....(1)

For Juliet, assuming that she is at rest at `t=0`
`s_J=6+0xxt+1/2at^2`
`s_J=6+1/2at^2` ......(2)

Under the given condition for them to meet
`s_R=s_J`
Equating (1) and (2) we get
`5t=6+1/2at^2`
`=>at^2-10t+12=0`
`=>a=(10t-12)/t^2` .....(3)

To find `a_max` we need to find first differential of `a(t)` with respect to `t` and set it equal to zero.

`(da)/dt=d/dt(10t-12)t^-2`

Using chain rule
`(da)/dt=(10t-12)(-2)t^-3+t^-2xx10`

Setting it equal `0` we get
`(10t-12)(-2)/t^3+10/t^2=0`
`=>(-10t+12)/t^3+5/t^2=0`
`=>(-10t+12)+5t=0`
`=>5t=12`
`=>t=12/5=2.4" s"`

Inserting this value in (3) we get
`a_max=(10xx2.4-12)/(2.4)^2`
`a_max=(10xx2.4-12)/(2.4)^2=2.08" ms"^-2`, rounded to two decimal places.