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A bus slows down uniformly from 75.0 km/h (21 m/s) to 0.0 km/h in 220 m. How long does it take to stop?

To find the time it takes for the bus to stop, we can use the formula for uniformly accelerated motion:

v^2 = u^2 + 2ax

Where:
v is the final velocity (0.0 km/h)
u is the initial velocity (75.0 km/h or 21 m/s)
a is the acceleration (uniform deceleration)
x is the distance covered (220 m)

First, let's convert the velocities to m/s:

Initial velocity (u) = 21 m/s
Final velocity (v) = 0.0 km/h = 0 m/s

Now, let's rearrange the formula to solve for the deceleration (a):

a = (v^2 - u^2) / (2x)

Substituting the values:

a = (0 - 21^2) / (2 * 220)

a = -441 / 440

a ≈ -1 m/s^2 (note the negative sign indicating deceleration)

Now, we can use the formula for uniformly accelerated motion to find the time (t):

v = u + at

0 = 21 + (-1)t

Solving for t:

-1t = -21

t = 21 seconds

Therefore, it takes 21 seconds for the bus to stop.

To determine how long it takes for the bus to stop, we can use the kinematic equation:

\(v_f = v_i + at\)

Where:
\(v_f\) is the final velocity (0.0 km/h or 0 m/s, since the bus has stopped)
\(v_i\) is the initial velocity (75.0 km/h or 21 m/s, as given in the question)
\(a\) is the acceleration
\(t\) is the time taken to stop

Since the bus slows down uniformly, we can assume its acceleration is constant throughout the stopping process. We can find the value of acceleration using another kinematic equation:

\(v_f^2 = v_i^2 + 2ad\)

Where:
\(d\) is the distance traveled (220 m, as given in the question)

We can rearrange the equation to solve for \(a\):

\(a = \frac{{v_f^2 - v_i^2}}{{2d}}\)

Plugging in the values:

\(a = \frac{{(0.0 \, \text{m/s})^2 - (21 \, \text{m/s})^2}}{{2 \times 220 \, \text{m}}}\)

Simplifying the calculation:

\(a = \frac{{-441}}{{440}} \, \text{m/s}^2\)

Now that we have the value of acceleration, we can use the first kinematic equation to solve for \(t\):

\(0.0 \, \text{m/s} = 21 \, \text{m/s} + \left(\frac{{-441}}{{440}} \, \text{m/s}^2\right)t\)

Simplifying the equation:

\(t = \frac{{-21 \, \text{m/s}}}{{-441/440}}\)

Calculating the result:

\(t \approx 0.952 \, \text{s}\)

Therefore, it takes approximately 0.952 seconds for the bus to stop.