An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Find the total time required for the police car to over take the automobile.
Answer:
15.02 sec
Explanation:
The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.
So; we can say :
[tex]D_{pursuit} =D_{police}[/tex]
By using the second equation of motion to find the distance S;
[tex]S= ut + \dfrac{1}{2}at^2[/tex]
[tex]D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)[/tex]
[tex]D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)[/tex]
[tex]D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)[/tex]
[tex]D_{police} = ut _P + \dfrac{1}{2}at_p^2[/tex]
where ;
u = 0
[tex]D_{police} = \dfrac{1}{2}at_p^2[/tex]
[tex]D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2[/tex]
[tex]D_{police} = 0.98*(t+12)^2[/tex]
[tex]D_{police} = 0.98*(t^2 + 144 + 24t)[/tex]
[tex]D_{police} = 0.98t^2 + 141.12 + 23.52t[/tex]
Recall that:
[tex]D_{pursuit} =D_{police}[/tex]
[tex](187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t[/tex]
[tex](187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0[/tex]
= 46.68 - 7.85 t -2.505 t² = 0
Solving by using quadratic equation;
t = -6.16 OR t = 3.02
Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs
From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;
Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s
Total time required for the police car to over take the automobile = 15.02 sec
