I am stuck on both of these questions. Would you please tell me if my reasoning is correct?
1. An object is dropped off a building. When comparing the average velocity of the object over equal time intervals, explain the difference between early time intervals and later time intervals
Would this be because the object would gain more speed as it gets closer to the ground?
2. the velocity of an object is given by the function v(t)=2t^3-t^2-15t. At which of the following times t is the object at rest?
a) 1
b)-2.5
c)3
d)5
I tried to solve this question by substituting each option into the equation to see which number would make v(t) equal zero. This did not work for me since b and c both made the equation equal zero.
#1 yes. the speed has a constant rate of increase
#2
v(t) = t(2t^2-t-15) = t(2t+5)(t-3)
You are correct, that b and c are both solutions. But doesn't negative time present a problem? I think that in the real world I'd have to go with just (c).
Let's go through each question and analyze your reasoning.
1. An object is dropped off a building. When comparing the average velocity of the object over equal time intervals, explain the difference between early time intervals and later time intervals.
Your reasoning is partially correct. As the object falls, it gains speed due to gravity, which results in an increase in velocity. This means that during later time intervals, the average velocity will be greater than during earlier time intervals.
However, it's important to note that the term "average velocity" refers to the total displacement divided by the total time. Since the object is dropped from rest, the total displacement will be negative (downward) regardless of the time interval considered. But as time goes on, the object covers a greater distance, resulting in a larger negative displacement and thus a larger average velocity.
2. The velocity of an object is given by the function v(t) = 2t^3 - t^2 - 15t. At which of the following times t is the object at rest?
Your initial approach of substituting each option into the equation to find when v(t) is equal to zero is on the right track. However, it seems that you encountered a problem when both options b) and c) made the equation equal zero.
Let's try finding the correct answer by solving the equation v(t) = 0 algebraically:
2t^3 - t^2 - 15t = 0
First, factor out a common term:
t(2t^2 - t - 15) = 0
Now, we can apply the zero-product property, which states that if a product of terms equals zero, then at least one of the terms must be equal to zero. In this case, we have:
t = 0 (from t = 0)
2t^2 - t - 15 = 0 (from 2t^2 - t - 15 = 0)
To solve the quadratic equation 2t^2 - t - 15 = 0, we can factor it or use the quadratic formula. Factoring doesn't yield nice whole number roots here, so let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 2, b = -1, and c = -15. Substituting these values into the formula, we get:
t = (1 ± sqrt((-1)^2 - 4(2)(-15))) / (2(2))
t = (1 ± sqrt(1 + 120)) / 4
t = (1 ± sqrt(121)) / 4
t = (1 ± 11) / 4
This simplifies to two potential solutions:
t = 12/4 = 3
t = -10/4 = -2.5
So, the correct answer is both t = 3 and t = -2.5. This means that at those times, the object will be at rest.