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

Question

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.

2985 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-17T21:22:02

This is what I get

(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.