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A driver is traveling at 80 km/h and comes to a stop in 3 seconds. How far does the car travel before coming to a stop.-O- Vf T=3s VF=0 Vi=80km/h

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Final answer:

The car, initially traveling at 80 km/h, converts to 22.22 m/s, and with a deceleration over 3 seconds, it travels 33.33 meters before coming to a stop.

Step-by-step explanation:

The student has asked how far a car will travel before coming to a stop if it is initially traveling at 80 km/h and comes to a stop in 3 seconds. To find the distance, we need to first convert the initial velocity to meters per second (m/s), since our answer will need to be in meters.

To convert from km/h to m/s, we multiply the speed in km/h by 1000 to get meters, and then divide by 3600 to get seconds. So, 80 km/h is equivalent to approximately 22.22 m/s (80 * 1000 / 3600).

Next, since the final velocity (Vf = 0 m/s) and the time (t = 3 s) are given, we can use the equation for deceleration (which is acceleration in the negative direction) to find the distance traveled:

d = Vi * t + (1/2) * a * t^2.

Here, Vi is the initial velocity and a is the acceleration.

To find acceleration, we use: a = (Vf - Vi) / t

Which gives us: (0 m/s - 22.22 m/s) / 3 s = -7.407 m/s^2.

Now we calculate the distance: (22.22 m/s * 3 s) + (1/2 * -7.407 m/s^2 * (3 s)^2), which results in 33.33 meters.

Therefore, the distance the car travels before coming to a stop is 33.33 meters.

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