The maximum retardation of a lift is 3m/s squared. Calculate the shortest time required to bring it to rest from a speed of 9m/s. What is the distance travelled in this timr
what is vo and a?
To calculate the shortest time required to bring the lift to rest, we can use the formula:
\(t = \frac{v}{a}\)
where:
- \(t\) is the time
- \(v\) is the initial velocity
- \(a\) is the retardation (acceleration in the opposite direction)
In this case, the initial velocity \(v\) is 9 m/s, and the retardation \(a\) is 3 m/s².
Substituting these values into the formula, we have:
\(t = \frac{9 \, \text{m/s}}{3 \, \text{m/s²}}\)
Simplifying the equation, we get:
\(t = 3 \, \text{s}\)
Therefore, the shortest time required to bring the lift to rest is 3 seconds.
To calculate the distance traveled in this time, we can use the equation:
\(d = v \cdot t\)
where:
- \(d\) is the distance
- \(v\) is the initial velocity
- \(t\) is the time
Substituting the values from the problem statement, we have:
\(d = 9 \, \text{m/s} \cdot 3 \, \text{s}\)
Simplifying the equation, we get:
\(d = 27 \, \text{m}\)
Therefore, the distance traveled in this time is 27 meters.
a. V = Vo + g*t = 0.
9 + (-3)t = 0,
t = 3 s.
b. V^2 = Vo^2 + 2a*d = 0.
9^2 + (-6)d = 0,
d = 13.5 m.