A bus slows down uniformly from 50.9 km/h (14.1 m/s) to 0 km/h in 27 s. How far does it travel before stopping?
v=v₀ -at
v=0 => t= v₀/a
s=v₀t -at²/’2
can somebody help me solve this?
To find the distance traveled by the bus before stopping, we can use the equation of motion:
\[d = \frac{{(v_i + v_f)}}{2} \cdot t\]
where:
- \(d\) is the distance traveled
- \(v_i\) is the initial velocity
- \(v_f\) is the final velocity
- \(t\) is the time taken
Given that the initial velocity (\(v_i\)) is 50.9 km/h (or 14.1 m/s), the final velocity (\(v_f\)) is 0 km/h (or 0 m/s), and the time taken (\(t\)) is 27 s, we can substitute these values into the equation:
\[d = \frac{{(14.1 + 0)}}{2} \cdot 27\]
Now, we can calculate the distance traveled:
\[d = 7.05 \cdot 27\]
Simplifying further:
\[d = 190.35 \, \text{m}\]
Therefore, the bus travels a distance of 190.35 meters before stopping.