Answer:18.55 s
Explanation:
Given
Police car velocity[tex]=80 km/h\approx 38.889 m/s[/tex]
speeder velocity[tex]=140 km/h\approx 22.22 m/s[/tex]
After 2.5 s Police car steps on accelerator to produce an acceleration of [tex]2.40 m/s^2[/tex]
In 5 sec speeder has traveled an additional distance of
=(38.889-22.22)2.5
=41.67 m
Therefore Police needs to travel a distance of 41.67 + distance traveled by speeder in next t sec
Distance traveled by speeder is s
[tex]s=38.889\times t[/tex]---1
By police car
[tex]s+41.67=22.22\times t+\frac{2.4\times t^2}{2}[/tex]---2
Subtract 1 from 2
[tex]41.67=22.22t+1.2t^2-38.889t[/tex]
[tex]1.2t^2-16.67-41.67=0[/tex]
t=16.05 s
Thus a total of 2.5+16.054=18.55 s
