An object starts from rest at the origin of an xy-coordinate system and has an acceleration with components ax=a and ay=a. After t seconds, the displacement and velocity of the object are Δd1 and v1, respectively. The object is stopped and returned to the origin. The object then starts at rest again with acceleration components ax=2a and ay=2a. After t seconds, the displacement and velocity of the object are Δd2 and v2, respectively. Which of the following correctly relates Δd1 and v1 to Δd2 and v2, respectively?
Responses
Δd2=4Δd1 and v2=2v1
Δd2=2Δd1 and v2=2√v1
Δd2=2√Δd1 and v2=2√v1
Δd2=5√Δd1 and v2=5√v1
Δd2=2Δd1 and v2=2v1
Hmm, let me calculate this for you using my trusty clown calculator. *starts juggling*
Wait a second, I just realized that this is a physics question, not a joke! Silly me!
Alright, let's see. The components of acceleration in the second scenario are double that of the first scenario. So, the displacement in the second scenario would be doubled compared to the first one. That means, Δd2=2Δd1.
As for the velocity, in the second scenario the acceleration is double, which means the change in velocity would also be double compared to the first scenario. Therefore, v2=2v1.
So, the correct answer is Δd2=2Δd1 and v2=2v1.
Now, that's some clown physics for you!
To determine the correct relationship between Δd1, v1, Δd2, and v2, let's analyze the problem step by step.
First, let's find the displacement (Δd1) and velocity (v1) after time t using the given acceleration components ax = a and ay = a.
We know that acceleration (a) is the derivative of velocity (v) with respect to time (t):
a = dv/dt
Integrating both sides with respect to time gives us:
v = at + C1
Since the object starts from rest, we have v=0 when t=0:
0 = a(0) + C1
C1 = 0
Therefore, the velocity v1 after time t is:
v1 = at
Next, let's calculate the displacement (Δd1) after time t using the velocity-time equation:
v = ds/dt
Integrating both sides with respect to time gives us:
s = vt + C2
Since the object starts from rest, we have s=0 when t=0:
0 = a(0)t + C2
C2 = 0
Therefore, the displacement Δd1 after time t is:
Δd1 = vt = at^2
Now, let's find the displacement (Δd2) and velocity (v2) after time t using the given acceleration components ax = 2a and ay = 2a.
Using the same approach as before, the velocity v2 after time t is:
v2 = (2a)t = 2at
And the displacement Δd2 after time t is:
Δd2 = (2at)(t) = 2at^2
Now we can compare the relationships between Δd1, v1, Δd2, and v2.
Δd2 = 4Δd1 and v2 = 2v1
Comparing the expressions we derived:
Δd2 = 2at^2 and Δd1 = at^2
v2 = 2at and v1 = at
Therefore, the correct relationship is:
Δd2 = 2Δd1 and v2 = 2v1
So the correct option is:
Δd2 = 2Δd1 and v2 = 2v1.
The correct relationship between Δd1 and Δd2 is given by:
Δd2 = 2Δd1
This means that the displacement after the second time period (t) is twice the displacement after the first time period (t).
The correct relationship between v1 and v2 is given by:
v2 = 2v1
This means that the velocity after the second time period (t) is twice the velocity after the first time period (t).
Therefore, the correct answer is: Δd2=2Δd1 and v2=2v1.