Answer:
Timing of catch up = 9.45 seconds after acceleration of police car.
Explanation:
Speed of police car = 95 km/h = 95 x 5 / 18 = 26.39 m/s
Speed of speeder = 135 km/h = 135 x 5 / 18 = 37.5 m/s
Let the time of catch up be t seconds.
Distance travelled by police car must be equal to distance travelled by speeder.
Distance travelled by speeder = Speed x Time = 37.5 x ( t + 1)
Distance travelled by police car is found out by [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time after 1 second.
In this case the velocity of police car in vertical direction = 26.39 m/s, acceleration = 2.60 [tex]m/s^2[/tex] and time is t.
Distance travelled by police car
[tex]s= 26.39*1+26.39t+\frac{1}{2} *2.6*t^2=26.39(t+1)+1.3t^2[/tex]
Equating both distances,
[tex]26.39(t+1)+1.3t^2=37.5(t+1)\\ \\ 11.11(t+1)=1.3t^2\\ \\ 1.3t^2-11.11t-11.11=0\\ \\ t=9.45s\\ \\ t=-0.90s[/tex]
Negative time is not possible,
So time is t = 9.45 seconds
Timing of catch up = 9.45 seconds after acceleration of police car.
