A train is traveling south at 23.5 m/s when the brakes are applied. It slows down with a constant rate to a speed of 5.01 m/s in a time of 8.73 s. How far does the train travel during the 8.73 s?

I have already found
-Acceleration during braking: -2.12m/s

vf^2=vi^2+2ad

put in your vf, vi, and a solve for d.

Thank you! I was stuck on this problem

So,

vf = final velocity = 5.01m/s
vi = initial velocity = 23.5m/s
a = acceleration = -2.12m/s^2
d = distance

I plugged everything in and got 0.045.
Is this the correct way to solve?

5.01^2 = 23.5^2 + 2(-2.12)d
25.1001 = 548.01d
d = 25.1001/548.01

Thanks!

I didn't check your math, but that is the way, you are not paying attenion to sig figures.

To find the distance the train travels during the 8.73 s, we can use the equation of motion:

\[ \text{Final velocity}^2 = \text{Initial velocity}^2 + 2 \times \text{acceleration} \times \text{distance} \]

Given:
Initial velocity (\(V_0\)) = 23.5 m/s
Final velocity (\(V\)) = 5.01 m/s
Acceleration (\(a\)) = -2.12 m/s (negative because it's deceleration)
Time (\(t\)) = 8.73 s

Rearranging the equation, we can solve for the distance (\(d\)):
\[ 2 \times \text{acceleration} \times \text{distance} = \text{Final velocity}^2 - \text{Initial velocity}^2 \]

Substituting the known values:
\[ 2 \times -2.12 \times d = 5.01^2 - 23.5^2 \]

Now, calculate the difference of squares on the right-hand side of the equation:
\[ 2 \times -2.12 \times d = 25.1001 - 552.25 \]

Simplifying the equation:
\[ -4.24 \times d = -527.1499 \]

Dividing both sides of the equation by -4.24:
\[ d = \frac{-527.1499}{-4.24} \]

Finally, compute \(d\):
\[ d \approx 124.441\, \text{m} \]

Therefore, the train travels approximately 124.44 meters during the 8.73 seconds.