A bus slows down uniformly from 75 km/h to 15 km/h in 12.55 s. How far did the car travel before stopping?

a=(V-Vo)/t = (15-75)/12.55 = -4.78 m/s^2

d = Vo*t + 0.5*a*t^2
d = 75*12.55 + 0.5*(-4.78)*12.55^2 =
564.8 m.

To find the distance traveled by the bus before stopping, we can use the formula for uniformly accelerated motion which relates acceleration (a), initial velocity (u), final velocity (v), and time (t):

\[v = u + at\]

Rearranging the equation, we get:

\[a = \frac{{v - u}}{{t}}\]

In this case, the initial velocity (u) of the bus is 75 km/h, the final velocity (v) is 15 km/h, and the time (t) is 12.55 s. We need to convert the velocities from km/h to m/s to ensure all units are consistent:

Initial velocity (u) = 75 km/h = \(75 \times \frac{1000}{3600}\) m/s

Final velocity (v) = 15 km/h = \(15 \times \frac{1000}{3600}\) m/s

Substituting the values into the equation gives:

\[a = \frac{{(15 \times \frac{1000}{3600}) - (75 \times \frac{1000}{3600})}}{{12.55}}\]

Simplifying further:

\[a = \frac{{(15 \times 1000) - (75 \times 1000)}}{{12.55 \times 3600}}\]

Now, we calculate the acceleration (a):

\[a = \frac{{15000 - 75000}}{{12.55 \times 3600}}\]

\[a = \frac{{-60000}}{{12.55 \times 3600}}\]

Next, we can use the following kinematic equation to find the distance (s) traveled by the bus before stopping:

\[s = ut + \frac{1}{2}at^2\]

Since the bus comes to a stop (final velocity v = 0), the equation simplifies to:

\[s = ut\]

Substituting the values:

\[s = (75 \times \frac{1000}{3600}) \times 12.55\]

Simplifying further:

\[s = (\frac{75 \times 1000}{3600}) \times 12.55\]

Calculating the distance traveled by the bus:

\[s = (\frac{75000}{3600}) \times 12.55\]

\[s = (\frac{75000}{3600}) \times 12.55\]

\[s = 262.5\]

Therefore, the bus traveled a distance of 262.5 meters before stopping.

To find the distance traveled by the bus before stopping, we need to use the equation for uniformly accelerated motion:

\(d = \frac{1}{2} (v_f + v_i) t\)

where:
\(d\) is the distance traveled,
\(v_f\) is the final velocity,
\(v_i\) is the initial velocity, and
\(t\) is the time taken.

Given:
\(v_i\) = 75 km/h,
\(v_f\) = 15 km/h, and
\(t\) = 12.55 s.

Before we can calculate the distance, we need to convert the velocities from km/h to m/s.

To convert km/h to m/s, we divide the value by 3.6 since there are 3.6 seconds in 1 hour.

So, \(v_i\) = 75 km/h ÷ 3.6 = 20.83 m/s
and \(v_f\) = 15 km/h ÷ 3.6 = 4.17 m/s

Now, we can substitute the values into the equation:

\(d = \frac{1}{2} (4.17 + 20.83) \times 12.55\)

Calculating the value inside the parentheses:

\(d = \frac{1}{2} (25) \times 12.55\)

\(d = 12.5 \times 12.55\)

\(d = 156.875\) meters

Therefore, the car traveled approximately 156.875 meters before stopping.