Question

1. -/1 points a car starts at a position of x = 9 mi and has a speed of 3 mph. after 39 s the car has a speed of 21 mph. determine the magnitude of the cars average acceleration during this time. mph/s 5. -/1 points a rock is dropped from rest. how fast is it going after it has been falling for 1.4 s? m/s

Explanation:

Problem 4

Step1: Recall the formula for average acceleration

The formula for average acceleration \(a_{avg}=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t\) is the time interval.
Given \(v_i = 3\) mph, \(v_f=21\) mph, and \(\Delta t = 39\) s.

Step2: Calculate the change in velocity

\(\Delta v=v_f - v_i=21 - 3=18\) mph.

Step3: Calculate the average acceleration

\(a_{avg}=\frac{\Delta v}{\Delta t}=\frac{18}{39}\approx0.46\) mph/s.

Step1: Recall the formula for the velocity of a falling object

For an object in free - fall (dropped from rest, \(v_0 = 0\)), the velocity formula is \(v = v_0+gt\). In the English system, \(g = 32\) ft/s². First, we need to convert the units if necessary. But if we assume the formula \(v = gt\) (since \(v_0 = 0\)).
Given \(t = 1.4\) s and \(g = 32\) ft/s².

Step2: Calculate the velocity

\(v=gt=32\times1.4 = 44.8\) ft/s.

Answer:

\(0.46\) mph/s

Problem 5