Question

question 14 (1 point)
a car is traveling south for 250 meters over the course of 100 seconds. what was the
cars average velocity?
a) 0.4 m/s south b) 0.4 m/s
c) 2.5 m/s d) 2.5 m/s south
question 15 (1 point)
if you were to find the area under a velocity - time graph, which of the following
quantities would you be calculating?
(a) displacement (b) instantaneous velocity
(c) average acceleration (d) time

Explanation:

Question 14

Step1: Recall the formula for average velocity

Average velocity \(v_{avg}=\frac{\Delta x}{\Delta t}\), where \(\Delta x\) is displacement and \(\Delta t\) is time.

Step2: Substitute the given values

Given \(\Delta x = 250\space m\) (south - direction is important for velocity) and \(\Delta t=100\space s\). Then \(v_{avg}=\frac{250}{100}=2.5\space m/s\). Since displacement is south - ward, velocity is \(2.5\space m/s\) south.

The area under a velocity - time graph is calculated as \(A=\int_{t_1}^{t_2}v(t)dt\). By the fundamental theorem of calculus, \(\int_{t_1}^{t_2}v(t)dt=x(t_2)-x(t_1)\), where \(x(t)\) is the position function. So, the area under the velocity - time graph gives the displacement.

Answer:

d) \(2.5\space m/s\) south

Question 15