Precisely 1.50 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20 m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)

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

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

there are no contact forces mentioned on the list of choices that you've provided.

according to jean piaget, in what stage of cognitive development are children able to understand conservation task?

b. preoperational

just took it