Question

practice problems:

1. youre driving and approaching a stop sign. you apply the brakes for 2.5s and the velocity of the car decreases to 4.75m/s. if the cars decelerates at a rate of 2.5m/s², what is the cars displacement during this time interval?
2. a ball is dropped from a balcony 45m above the ground. how long does it take the ball to hit the ground?

Explanation:

Step1: Find the initial velocity

Use the formula \(v = v_0+at\). Rearranging for \(v_0\) gives \(v_0=v - at\).
Given \(v = 4.75m/s\), \(a=- 2.5m/s^{2}\), \(t = 2.5s\)
\(v_0=4.75-(-2.5)\times2.5=4.75 + 6.25=11m/s\)

Step2: Calculate the displacement

Use the formula \(x=v_0t+\frac{1}{2}at^{2}\)
Substitute \(v_0 = 11m/s\), \(a=-2.5m/s^{2}\), \(t = 2.5s\)
\(x=(11\times2.5)+\frac{1}{2}\times(-2.5)\times(2.5)^{2}\)
\(x = 27.5-7.8125\)
\(x=19.6875m\)

Answer:

\(19.6875m\)