A police cruiser is moving at a constant speed of 90 km/h when it is suddenly passed by a speeder moving at 145 km/h. Precisely two seconds later, the police officer steps on the gas and accelerates at a rate of 2.30 m/s^2m/s 2 . How long does it take for the police car to reach the position of the speeder? (Assume the speeder continues to move at a constant speed). Use 's' for your units.

Explanation:

Given that,

The police car is moving at a constant speed

V=90km/hr =90×1000/3600=25m/s

Since the car is moving at constant speed then, it acceleration is zero.

Another car who is at a high speed passed the police at a speed of

V' = 145km/hr = 40.28m/s

Speeder is moving at constant speed, then, a=0m/s²

After 2 seconds, the police car accelerate to 2.3m/s²

a = 2.3m/s²

Let assumed that the police car meets the speeder at d

Note, at the first 2 seconds,

The speeder will be at

d = ut + ½at², a=0

d = 40.28×2

d = 80.56m

And the police car will be at

d =ut = 25×2

d = 50m

So the speeder is already ahead at a distance of 80.56—50= 30.56m

So, let analyse from this new point and add the 2sec later,

Let assume the speeder traveled 'd' distance before they meet

Then, the police car will have to travel d+30.56 distance

Then, let calculate the time to reach distance, because they will spent the same time to reached there

For the speeder

d = ut

Then, d = 40.28t

For the police

d+30.56 = ut + ½at²

d = ut+½at² —30.56

d = 25t + ½•2.3•t²—30.56

d = 25t + 1.15t² - 30.56

So equating the two distances.

25t + 1.15t² - 30.56 = 40.28t

1.15t² —15.28t - 30.56 = 0

Divide through by 1.15

t² —13.29t—26.57 = 0

Using formula method to solve the quadratic equation

t = [-b±√(b²-4ac)]/2a

Where a = 1, b = -13.29 and c=-26.57

Solving this and discarding the negative time, we have

t = 15.05 s

Then, the total time the police will use to get to the speedy car is

t(total) = 15.05+2

t = 17.05 seconds