Aspeeder passes a parked police car at 26.8 m/s. instantaneously, the police car starts from rest with a uniform acceleration of 2.40 m/s2 . how much time pases before the speeder is overtaken by the police car? answer in units of s.

The velocity of the speeder is expressed by the equation:

v1 = d1 / t1 --> d1 = v1 t1

And for the police:

d2= v2i t2 + 0.5 a2 t2^2

where v is final velocity, vi is initial velocity, d is distance, a is acceleration and t is time

The distance of the two must be equal when the police overtakes the speeder hence:

v1 t1 = v2i t2 + 0.5 a2 t2^2

t1 = t2 = t and v2i = 0 since the police starts at rest, so:

26.8 t = 0.5 * 2.4 * t^2

26.8 = 1.2 t

t = 22.33 seconds

im sorry i dont know

it has to be b, about 97 percent.