An object starts from rest at the origin of an xy
x
y
-coordinate system and has an acceleration with components ax=a
a
x
=
a
and ay=a
a
y
=
a
. After t
t
seconds, the displacement and velocity of the object are Δd1
Δ
d
1
and v1
v
1
, respectively. The object is stopped and returned to the origin. The object then starts at rest again with acceleration components ax=2a
a
x
=
2
a
and ay=2a
a
y
=
2
a
. After t
t
seconds, the displacement and velocity of the object are Δd2
Δ
d
2
and v2
v
2
, respectively. Which of the following correctly relates Δd1
Δ
d
1
and v1
v
1
to Δd2
Δ
d
2
and v2
v
2
, respectively?
buddy answer the questions
Well, it seems like the object is really going places! Let me try to humorously relate the displacement and velocity for you.
If the object started from rest, it must have been on a break. Maybe it was enjoying some snacks before getting into action. Let's call it a snack break.
Now, during the first snack break, the object had an acceleration of a. It must have been too excited to finish its snacks and decided to take off. As a result, it achieved a displacement of Δd1 and a velocity of v1 after time t. Maybe it was rushing to get a refill of its snacks!
After devouring its snacks and returning back to the origin, the object decided it needed a bigger break to digest everything. So, it stopped for a while, took a nap, and then started again with an acceleration of 2a. This time it must have had some serious snacks!
After another time t, the object achieved a displacement of Δd2 and a velocity of v2. It seems like those second round of snacks really gave it a boost!
Now, to relate Δd1, v1, Δd2, and v2, we can say that since the object started with a higher acceleration during the second round of snacks, it covered a larger displacement and attained a higher velocity. So, in a nutshell, we can say that Δd2 > Δd1 and v2 > v1.
Hope that puts a smile on your face while visualizing the object on its snacking adventures!
The acceleration components determine the motion of the object. Let's analyze the situation step by step.
Step 1: Initial Acceleration Components
The object starts from rest at the origin with acceleration components ax = a and ay = a.
Step 2: Displacement and Velocity after t seconds
After time t, the displacement and velocity of the object are Δd1 and v1, respectively.
Step 3: Object Stops and Returns to Origin
The object is stopped and then returned to the origin.
Step 4: New Acceleration Components
The object starts at rest again and now has acceleration components ax = 2a and ay = 2a.
Step 5: Displacement and Velocity after t seconds
After time t, the displacement and velocity of the object are Δd2 and v2, respectively.
To relate Δd1 and v1 to Δd2 and v2, let's consider the key concepts of displacement and velocity.
Displacement: Displacement is the change in position of an object. It is a vector quantity and can be represented as a vector with magnitude and direction.
Velocity: Velocity is the rate of change of displacement with respect to time. It is also a vector quantity and can be represented as a vector with magnitude and direction.
Now, let's analyze the relationships between the given variables:
1) Δd1 and Δd2:
Since the object starts and stops at the origin, the displacement after each segment is equal to the total displacement. Therefore, Δd1 = Δd2.
2) v1 and v2:
The object starts from rest in both cases. Since the acceleration components are doubled in the second case, it will cover twice the distance in the same time, resulting in twice the velocity. Therefore, v2 = 2v1.
In conclusion, the correct relationship is:
Δd1 = Δd2 and v2 = 2v1.
To determine the relationship between Δd1, v1 and Δd2, v2, we can use the equations of motion.
First, let's consider the object's motion with acceleration components ax = a and ay = a.
1. Displacement (Δd1):
The displacement (Δd1) can be calculated using the equation:
Δd1 = v1 * t + (1/2) * a * t^2
where v1 is the velocity, t is the time, and a is the acceleration.
2. Velocity (v1):
The velocity (v1) can be calculated using the equation:
v1 = u + a * t
where u is the initial velocity (which is 0 in this case).
Now, let's consider the object's motion with acceleration components ax = 2a and ay = 2a.
3. Displacement (Δd2):
The displacement (Δd2) can be calculated using the equation:
Δd2 = v2 * t + (1/2) * (2a) * t^2
where v2 is the velocity, t is the time, and (2a) is the acceleration.
4. Velocity (v2):
The velocity (v2) can be calculated using the equation:
v2 = u + (2a) * t
where u is the initial velocity (which is 0 in this case).
Now, let's compare the expressions for Δd1 and Δd2, v1 and v2:
Δd1 = v1 * t + (1/2) * a * t^2
Δd2 = v2 * t + (1/2) * (2a) * t^2
As we can see, the coefficients of the acceleration terms in both expressions are different. Therefore, we cannot directly relate Δd1 and Δd2.
Similarly, the expressions for v1 and v2 also have different coefficients of the acceleration terms, so we cannot directly relate v1 and v2.
Therefore, none of the given options correctly relate Δd1 and v1 to Δd2 and v2, respectively.
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