HELP PLEASEEEE>??

An unmarked police car travelling at a constant 80.0 km/h is passed by a speeder
travelling at 100.0 km/h. Precisely 1.0 s after the speeder passes, the police steps on the accelerator. If the police car’s acceleration is 2.0 m/s2, how much time passes before the police car overtakes the speeder (assumed moving at a constant speed)?

1 Answer
Mar 8, 2017

Answer:

#6.4s#, rounded to one decimal place.

Explanation:

Let the Police car overtake the speeder after a period #t# passes once he crosses the police car.

  1. For first #1.0s# both cars travel at their respective speeds. Distance between the two cars at the end of #1.0s#
    #d_1="Relative speed of cars" xx"1.0"#
    #d_1=[(100.0-80.0)xx1000/3600] xx"1.0"=20000.0/3600=50/9m#
    This is the distance speeder is ahead of police car just before it accelerates.
  2. Distance moved by speeder after #1.0s# before Police car overtakes it
    #d_2=100.0xx1000/3600xx(t-1)=250/9(t-1)m#
  3. Distance to be covered by Police car to overtake the speeder in #(t-1) s# is
    #d_1+d_2=50/9+250/9 (t-1)m#
  4. Setting up the kinematic expression
    #s=ut+1/2at^2#
    and inserting given values in SI units we get
    #50/9+250/9 (t-1)=200/9(t-1)+1/2xx2.0(t-1)^2#
    #=>(t-1)^2-50/9(t-1)-50/9=0#
    #=>9(t-1)^2-50(t-1)-50=0#
    To solve the quadratic we substitute #t-1=x#
    It becomes
    #9x^2-50x-50=0#

I found roots of this quadratic using inbuilt graphic tool
my comp

we get roots as #x=-0.9 and 6.4#, rounded to one decimal place
Time can not be negative, therefore, ignoring the first root we get

#=>t-1=6.4#
#=>t=6.4s#, rounded to one decimal place.