Question

1. figure is an acceleration - time graph for an object.

(a) calculate the change in velocity over the time interval 2.0 s to 5.0 s.
(b) if the starting velocity for the time interval in (a) is 6.0 m/s s, what is the final velocity of the object at the end of the time interval

Explanation:

Step1: Calculate the change in velocity

The change in velocity \(\Delta v\) is given by the area under the acceleration - time graph.
Since the acceleration \(a = 4.0\ m/s^{2}\) (constant) and the time interval \(\Delta t=5.0 - 2.0=3.0\ s\)
The formula for the area of a rectangle (since \(a\) is constant) is \(\Delta v=a\times\Delta t\)
\(\Delta v = 4.0\times(5.0 - 2.0)\)
\(\Delta v=4.0\times3.0\)
\(\Delta v = 12\ m/s\)

Step2: Calculate the final velocity

The formula for velocity is \(v = v_{0}+\Delta v\)
Given \(v_{0}=6.0\ m/s\) and \(\Delta v = 12\ m/s\)
\(v=6.0 + 12\)
\(v = 18\ m/s\)

Answer:

(a) The change in velocity is \(12\ m/s\)
(b) The final velocity is \(18\ m/s\)