Consider the data collected in science class. Different masses were thrown with varied amounts of
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Answer:
Explanation:
Assuming the squirrel is jumping off the ground, here's what we know but don't really know...
v₀ = 4.0 at 50.0°
So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:
which gives us that the upward velocity is
v₀ = 3.1 m/s
Moving on here's what we also know:
a = -9.8 m/s/s and
v = 0
Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:
v = v₀ + at and filling in:
0 = 3.1 - 9.8t and
-3.1 = -9.8t so
t = .32 seconds.
Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:
Δx = and filling in:
Δx = and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:
Δx = .99 - .50 so
Δx = .49 meters
The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.
v = v₀ + a t
y = v₀ t + ½ a t²
where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.
In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:
- Positive slope the body is accelerating and the speed is increasing.
- Negative slope the body is stopping, the speed decreases.
Let's answer the different questions about the system.
a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s
v = v₀ + a t
a = 0.25 m / s²
c) The maximum acceleration is when the slope is maximum.
a = 1 m / s²
Therefore the acceleration is maximum in the section between 10 and 14 s
d) The total displacement is the sum of the displacements of each section.
We look for every displacement.
d₁ = v₀ + ½ a₁ Δt²
d₁ = 0 + ½ 0.25 8²
d₁ = 8 m
In the second section the velocity is constant
d₂ = v₂ Δt₂
d₂ = 2 (10-8)
d₂ = 4 m
The third section.
d₃ = v₀ + ½ a t²
d₃ = 2 + ½ 1 (14-10) ²
d₃ = 10 m
The distance of the fourth section.
we look for acceleration
a₄ =
a₄ =
a₄ = -1.5 m / s²
d₄ = 6 + ½ (-1.5) (1814) ²
d₄ = -6 m
The total displacement is;
= 8 + 4 + 10 -6
= 16 m
e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to
v = 6 m / s
In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Learn more about kinematics here: brainly.com/question/24783036
