Answer:
The distance traveled is 0.037 mi
Explanation:
The equation for the position and velocity of an accelerated object is:
x = x0 + v0 * t + 1/2 * a * t²
v = v0 + a * t
where
x = position at time t
x0 = initial position
t = time
a = acceleration
v0 = initial velocity
If the velocity is constant, then a = 0 and the position will be:
x = x0 + v * t where "v" is the velocity
First, let´s find the distance traveled until the driver push the brake:
The speed is constant. Then:
x = x0 + v * t (considering the origin of the reference system to be located at the point at which the driver sees the puppy, x0 = 0)
x = 35 mi/h (1 h / 3600 s) * 1.7 s = 0.017 mi
Then, the drivers moves with constant acceleration until the car stops (v = 0)
From the equation for velocity:
v = v0 + a * t
Since v = 0, we can obtain the acceleration of the car until it stops. With that acceleration, we can calculate how much distance the car moves before it stops.
0 = v0 + a * t
-v0 / t = a
-35 mi/h (1 h / 3600s) / 4.0 s = a
a = -2.4 x 10⁻³ mi/s²
The distance traveled will be:
x = x0 + v0 * t + 1/2 * a * t²
Now x0 will be the distance traveled before the driver slows down.
x = 0.017 mi + 35 mi/h (1 h / 3600s) * 4 s + 1/2 * ( -2.4 x 10⁻³ mi/s²) * (4s)²
x = 0.037 mi
