Answer:
2.58 seconds
Step-by-step explanation:
Given: the shuttle bus comes to a sudden stop to avoid hitting a dog, it decelerates uniformly at \(3.8\) \(\text{m}/\text{s}^2\) as it slows from \(9.8\) \(\text{m}/\text{s}\) to \(0\) \(\text{m}/\text{s}\).
To Find: the time interval of acceleration for the bus.
Solution:
Initial speed of shuttle bus \(\text{u}=9.8\) \(\text{m}/\text{s}\)
Final speed of shuttle bus \(\text{v}=0\) \(\text{m}/\text{s}\)
acceleration \(=3.8\) \(\text{m}/\text{s}^2\)
using first equation of motion
\(\text{v}=\text{u}+\text{a}\text{t}\)
as speed decreases acceleration will be negative
\(\text{v}=\text{u}-\text{a}\text{t}\)
putting value in equation
\(0=9.8-3.8\text{t}\)
\(3.8\text{t}=9.8\)
\(\text{t}=\frac{9.8}{3.8}\)
\(\text{t}=2.58\) \(\text{seconds}\)
It takes \(2.58\text{s}\) to decelerate to \(0\) \(\text{m}/\text{s}\)
