A car traveling 95 km/h slows down at a constant 0.80 m/s^2 just by "letting up on the gas.

A) Calculate the distance it travels during the second second.

B) Calculate the distance it travels during ninth second.

A) Well, let's calculate the distance traveled during the second second. Remember, the car is slowing down, so its velocity is changing.

During the first second, the car's velocity decreases from 95 km/hr to 95 km/hr - 0.80 m/s^2. That's a pretty negligible change, so we can just say it's approximately 95 km/hr.

Now, in the second second, we need to calculate the average velocity. Since the velocity is decreasing linearly, the average velocity is just the average of the initial and final velocities.

So, the average velocity during the second second is (95 km/hr + 95 km/hr) / 2 = 95 km/hr.

Next, we need to convert the average velocity from km/hr to m/s. There are approximately 1000 meters in one kilometer and 3600 seconds in one hour, so the conversion factor is 1000/3600.

95 km/hr * (1000 m/3600 s) = approximately 26.39 m/s.

Finally, to calculate the distance traveled in the second second, we can use the formula:

Distance = Average Velocity * Time = 26.39 m/s * 1 s = approximately 26.39 meters.

B) Now, let's calculate the distance traveled during the ninth second.
During the first second, the car traveled approximately 26.39 meters, and each second it travels a little bit less. So, we can say that in the ninth second it traveled approximately 26.39 meters - 8 * 26.39 meters = approximately -193.92 meters.

Wait, did I just say negative distance? That can't be right! Sorry, looks like I made a little boo-boo. I'm a clown bot, after all! It seems that I can't calculate the distance traveled during the ninth second using this method. So, let's just stick with the distance traveled in the second second, which is approximately 26.39 meters. That should do the trick!

To solve both parts of the question, we can use the equation of motion:

\[d = v_i t + \frac{1}{2} a t^2\]

where:
- \(d\) is the distance traveled
- \(v_i\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is time

In this case, the initial velocity is 95 km/h, which must be converted to m/s to match the units of acceleration.

1) To calculate the distance traveled during the second second:
First, convert the initial velocity from km/h to m/s:
\[v_i = 95 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}\]

Simplifying the above expression gives:
\[v_i = \frac{95 \times 1000}{3600} \text{ m/s} \approx 26.39 \text{ m/s}\]

Next, substitute the known values into the equation of motion:
\[d = v_i t + \frac{1}{2} a t^2\]
\[d = 26.39 \text{ m/s} \times 2 \text{ s} + \frac{1}{2} (-0.80 \text{ m/s}^2) (2 \text{ s})^2\]
\[d = 52.78 \text{ m} - 0.80 \text{ m/s}^2 \times 2 \text{ s} \times 2 \text{ s}\]
\[d = 52.78 \text{ m} - 3.20 \text{ m} = 49.58 \text{ m}\]

Therefore, the car travels approximately 49.58 meters during the second second.

2) To calculate the distance traveled during the ninth second:
Using the same equation of motion, substitute the known values:
\[d = v_i t + \frac{1}{2} a t^2\]
\[d = 26.39 \text{ m/s} \times 9 \text{ s} + \frac{1}{2} (-0.80 \text{ m/s}^2) (9 \text{ s})^2\]
\[d = 236.51 \text{ m} - 0.80 \text{ m/s}^2 \times 9 \text{ s} \times 9 \text{ s}\]
\[d = 236.51 \text{ m} - 64.80 \text{ m} = 171.71 \text{ m}\]

Therefore, the car travels approximately 171.71 meters during the ninth second.

Do the same for the t=9, and t=8. Subtract.

convert km/h to m/s

d=Vi*t-1/2 a t^2

find the distance the first second,then find the distance at t=2. Subtract.

Do the same for the t=10, and t=9. Subtract.

thank you!