Answer:
11.42 s
Explanation:
Assuming the speed of the police car = 99 Km/h
v = 99 km/h x 0.278 = 27.52 m/s
Speed of the Passer = 133 Km/h
u = 133 x 0.278 = 37 m/s
acceleration of the police car = 2.20 m/s²
time taken to overtake passer = ?
distance travel by the passer in t second.
x = u t
x = 37 t
Distance traveled by the police
[tex]x = ut+ \dfrac{1}{2}at^2[/tex]
[tex]x = 27.52 t+ 0.5\times 2.2\times (t-1.5)^2[/tex]
both the distance are same so,
37 t = 27.52 t +1.1 (t-1.5)²
8.62 t = t² + 2.25 - 3t
t² -11.62 t + 2.25 = 0
on solving the above equation we get
t = 11.42 s , 0.92 s
neglecting the second value because it is less than 1.5 s.
Time taken by the driver to overtake the passer is equal to 11.42 s
