To pass a slow vehicle within the length of the passing lane roger got up well over the speed limit . He then spotted a police cruiser coming around the corner and hit the brakes . If he braked at 6 m/s^2 , and it took him 1.50 seconds to get back to 40 m/s , what speed had he reached before spotting the police ?

The instantaneous speed of a particle moving along one straight line is

v(t) = ate^−4t,
where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a?

V = Vo + a*t = 40, Vo - 6*1.5 = 40, Vo = 49 m/s.

To find the initial speed at which Roger was traveling before he spotted the police, we can use the formula for acceleration:

acceleration = (final speed - initial speed) / time

Rearranging the formula, we have:

initial speed = final speed - (acceleration * time)

Given:
acceleration (a) = -6 m/s^2 (negative because braking implies deceleration)
final speed (v) = 40 m/s
time (t) = 1.50 seconds

Substituting the given values into the formula, we get:

initial speed = 40 m/s - (-6 m/s^2 * 1.50 seconds)
initial speed = 40 m/s + 9 m/s
initial speed = 49 m/s

Therefore, Roger had reached a speed of 49 m/s before he spotted the police.