# HELP PLEASEEEE>??

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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 carâ€™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 carâ€™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:

#### 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