Acar is cruising at a steady speed of 35 mph. suddenly, a cuddly puppy runs out into the road. the driver takes 1.7 seconds to react, and then slams on the brakes. the brakes slow the car down at constant acceleration for 4.0 seconds until the car stops. the puppy, unharmed, happily skips away. 1. what is the total distance traveled between when the driver rst sees the puppy to when the car stops

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

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this