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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)?
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
Explanation:
Let the Police car overtake the speeder after a period
- 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. - Distance moved by speeder after
#1.0s# before Police car overtakes it
#d_2=100.0xx1000/3600xx(t-1)=250/9(t-1)m# - 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# - 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
we get roots as
Time can not be negative, therefore, ignoring the first root we get
