A driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s² . She is driving at 20 m/s when suddenly she sees an obstacle in the road 50 m in front of her. What is the distance she passes after noticing the obstacle before fully stopping?

43.3 m

Explanation:

given :

the initial velocity of the car , u = 20m/s

The reaction time is 0.5s

the distance that the car will travel during the reaction time

= initial velocity x reaction time

= 20 m/s  x 0.5 s

= 10 m

Now we turn our attention to the stopping distance AFTER she reacts and steps on the brakes:

recall that one of the forms that the equations of motions can be expressed is:

v² = u² + 2as

where v = final velocity = 0 because the car comes to a stop

u = initial velocity = 20m/s

a = acceleration = -6.0 m/s²  (note negative sign because it is a deceleration)

s = distance traveled (we are asked to find this)

substituting the known values into the equation:

v² = u² + 2as

0² = 20² + 2(-6.0) s

0 = 400 - 12s

1.2s = 400

s = 400/12

s = 33.33 m

Total distance travelled

= distance during reacting time + distance while decelerating

= 10  + 33.3

= 43.3 m