A motorist travels at a constant speed of 34.0 m/s through a school zone, exceeding the posted speed limit. A policeman waits 7.0 s before giving chase at an acceleration of 3.7 m/s2.

Question

anonymous · Answer

Find the time required to catch the car from the instant the car passes the policeman.

anonymous · Answer

Find the distance required for the policeman to overtake the motorist.

anonymous · Answer

For Car  ( insert the values and get an equation)
s = ut
s = 34 t ------> (1) equation

For Policeman ( he starts 7 s later so it is (t-7))

s= ut + 1/2 at^2 ( u = 0) as he strts frm rest

$$s = \frac{1}{2} \times 3.7 \times (t-7)^{2}$$

$$2s = 3.7(t^{2}-14t+49)$$

$$2s = 3.7t^2 -51.8t +181.3$$

Now plug in equation {1} in to s
so 

$$2(34t) = 3.7t^{2}-51.8t+181.3$$

i think now u can solve and find time.

anonymous · Answer

did u get it ...

anonymous · Answer

no?

anonymous · Answer

ok which one u dont understnd?

anonymous · Answer

im at 3.7t^2 - 119.8t + 181.3 = 0 now im working on the quadratic equation but I keep getting it wrong please help

anonymous · Answer

u can either use the calculator or 
this formula

anonymous · Answer

i tried the calculator and it keeps saying error

anonymous · Answer

Use this...its accurate http://www.google.lk/imgres?q=quadratic+formula&um=1&hl=en&sa=N&biw=1280&bih=675&tbm=isch&tbnid=wOsI1N_cpFKysM:&imgrefurl=http://www.glogster.com/kylie1597/quadratic-formula/g-6mf9s8clhc23fa1opvsbua0&imgurl=http://im.glogster.com/media/4/30/58/45/30584538.jpg&w=549&h=215&ei=6JA9UM3JOMTnrAflp4GYCQ&zoom=1&iact=hc&vpx=328&vpy=359&dur=1323&hovh=140&hovw=359&tx=199&ty=60&sig=106954449408673256726&page=1&tbnh=76&tbnw=193&start=0&ndsp=18&ved=1t:429,r:7,s:0,i:132

anonymous · Answer

or insert the values in this http://www.math.com/students/calculators/source/quadratic.htm