Acar is driving at a speed of 59.0 km/h toward an intersection just as the light changes from green to yellow. the driver has a reaction time of 0.750 s and the magnitude of the braking acceleration of the car is 5.50 m/s2. find the minimum distance xmin the car travels before coming to a stop after the light changes. (express to three sig figs)

The car travels 36.8 m before coming to stop after the light changes

Explanation:

The car moves at a constant speed of 59.0 km/h for 0.750 s before the driver hits the brake.

The equation for the position of an object moving at constant speed is:

x = x0 + v t

where:

x = position at time t

x0 = initial position

v = speed

t= time

Let´s consider the initial position as the position at which the driver sees the traffic light and decides to brake. That will make x0 = 0. Then, the position after 0.750 s will be:

x = 59.0 km/h * 0.750 s (1 h /3600 s) = 0.0123 km (1000 m / 1 km) = 12.3 m

while braking, the car has a negative acceleration, then, the speed is not constant. The position of the car will be given by the following equation:

x = x0 + v0 t + 1/2 a t² ( where a = acceleration and v0 = initial speed)

and the speed can be expressed as follows:

v = v0 + a t

from this equation, we can calculate how much time it takes the car to stop (v = 0):

0 = v0 + a t

-v0 = a t

-v0 / a = t

v0 is the speed of the car as the driver hits the brake (59.0 km/h) and "a" is the acceleration (5.50 m/s²) that will be negative because the car is losing speed. Then:

-59.0 km/h (1000 m / 1 km) (1 h / 3600 s) / (-5.50 m/s²) = 2.98 s

Now, we can calculate the position at this time to know the minimum distance the car travels before coming to stop:

x = x0 + v0 t + 1/2 a t²

now x0 will be the distance traveled after the driver sees the light but before braking ( 12.3 m).  v0 will be the speed before braking, 59.0 km / h or 16.4 m/s. Then:

x = 12.3 m + 16.4 m/s * 2.98 s +1/2 (-5.50 m/s²) * (2.98 s)²

x = 36.8 m

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