Can somebody help translate this question? 2D Kinematics.

In summary: When you want to find the magnitude of the acceleration the components add in quadrature (that is, square root of the sum of the squares).
  • #1
DavidAp
44
0
I'm having difficult understanding the question, can somebody help me? The question is:

"The velocity of a 3 kg particle is given by v = (3ti + 6(t^2)j )m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 35 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer:
(a) 75.099403312170 Units ° (degrees)
(b) 61.978933315223 Units ° (degrees)"


My initial approach was to find t when the magnitude of the particle is 35N to get the components of the triangle.

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3 + 12t

Time
F = ma
35 = 3(3+12t)
35 = 9 + 36t
26 = 36t
.72 = t

Since, I thought I was to find the angle created by the velocity vector, I found the y-comp. (opp.) and x-comp. (adj.).
y-comp. = 6(t^2) = 6(.72^2) = 3.13
x-comp. = 3t = 3(.72) = 2.27

tan(theta) = opp./adj. = 3.13/2.27
theta = arctan(3.13/2.27) = arctan(1.44) = 55.27

However, 55.27 isn't the angle... for anything! Did I misinterpret the question? What is the difference between the two angles they are asking me to find, what are they asking of my specifically? If somebody can help me I would deeply appreciate it.

Thank you for taking the time to read my question.
 
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  • #2
DavidAp said:
I'm having difficult understanding the question, can somebody help me? The question is:

"The velocity of a 3 kg particle is given by v = (3ti + 6(t^2)j )m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 35 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer:
(a) 75.099403312170 Units ° (degrees)
(b) 61.978933315223 Units ° (degrees)"


My initial approach was to find t when the magnitude of the particle is 35N to get the components of the triangle.

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3 + 12t

Time
F = ma
35 = 3(3+12t)
35 = 9 + 36t
26 = 36t
.72 = t

Since, I thought I was to find the angle created by the velocity vector, I found the y-comp. (opp.) and x-comp. (adj.).
y-comp. = 6(t^2) = 6(.72^2) = 3.13
x-comp. = 3t = 3(.72) = 2.27

tan(theta) = opp./adj. = 3.13/2.27
theta = arctan(3.13/2.27) = arctan(1.44) = 55.27

However, 55.27 isn't the angle... for anything! Did I misinterpret the question? What is the difference between the two angles they are asking me to find, what are they asking of my specifically? If somebody can help me I would deeply appreciate it.

Thank you for taking the time to read my question.

The velocity had an i and j component. 3ti + 6t2j

Should the acceleration also have had an i and j component.
 
  • #3
PeterO said:
The velocity had an i and j component. 3ti + 6t2j

Should the acceleration also have had an i and j component.
Yes... however, I'm still quite oblivious to how that will help :/. I just assumed that the i and j component represented the x and y component of the slope for the velocity. How will i and j help in the equation for acceleration?

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3i + 12tj

Time
F = ma
35 = 3(3i+12tj)
35 = 9i + 36tj
..? help?
 
  • #4
DavidAp said:
Yes... however, I'm still quite oblivious to how that will help :/. I just assumed that the i and j component represented the x and y component of the slope for the velocity. How will i and j help in the equation for acceleration?

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3i + 12tj

Time
F = ma
35 = 3(3i+12tj)
35 = 9i + 36tj
..? help?

because the directions i & j are perpendicular, I expect you would be working with Pythagorus.

(35/3)2 = 32 + (12t)2

That will let you find t, so then substitute into velocity and acceleration expressions to find the directions and magnitudes.
 
  • #5
DavidAp said:
Yes... however, I'm still quite oblivious to how that will help :/. I just assumed that the i and j component represented the x and y component of the slope for the velocity. How will i and j help in the equation for acceleration?

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3i + 12tj

Time
F = ma
35 = 3(3i+12tj)
35 = 9i + 36tj
..? help?

When you want to find the magnitude of the acceleration the components add in quadrature (that is, square root of the sum of the squares).
 
  • #6
gneill said:
When you want to find the magnitude of the acceleration the components add in quadrature (that is, square root of the sum of the squares).
Thank you so much! I really appreciate your help. :)
 
  • #7
DavidAp said:
I'm having difficult understanding the question, can somebody help me? The question is:

"The velocity of a 3 kg particle is given by v = (3ti + 6(t^2)j )m/s, with time t in seconds. At the instant the net force on the particle has a magnitude of 35 N, what are the direction (relative to the positive direction of the x axis) of (a) the net force and (b) the particle's direction of travel?

Answer:
(a) 75.099403312170 Units ° (degrees)
(b) 61.978933315223 Units ° (degrees)"


My initial approach was to find t when the magnitude of the particle is 35N to get the components of the triangle.

Velocity and Acceleration
v = 3ti + 6(t^2)j
a = 3 + 12t

Time
F = ma
35 = 3(3+12t)
35 = 9 + 36t
26 = 36t
.72 = t

Since, I thought I was to find the angle created by the velocity vector, I found the y-comp. (opp.) and x-comp. (adj.).
y-comp. = 6(t^2) = 6(.72^2) = 3.13
x-comp. = 3t = 3(.72) = 2.27

tan(theta) = opp./adj. = 3.13/2.27
theta = arctan(3.13/2.27) = arctan(1.44) = 55.27

However, 55.27 isn't the angle... for anything! Did I misinterpret the question? What is the difference between the two angles they are asking me to find, what are they asking of my specifically? If somebody can help me I would deeply appreciate it.

Thank you for taking the time to read my question.

So I solved a, but I don't know how to do part b. Anyone can help me out ?
 
  • #8
The direction of the velocity is the direction of travel.
 
  • #9
SammyS said:
The direction of the velocity is the direction of travel.

Thank you !
 

1. What is 2D kinematics?

2D kinematics is the study of motion in two-dimensional space, where an object's position, velocity, and acceleration are described in terms of both horizontal and vertical components.

2. Why is 2D kinematics important?

2D kinematics is important because it allows us to analyze and understand the motion of objects in real-world scenarios, such as projectiles, vehicles, and athletes. It also helps us make predictions and solve problems related to motion.

3. How do you solve problems involving 2D kinematics?

To solve problems involving 2D kinematics, you can use equations and principles from classical mechanics, such as Newton's laws of motion and the equations of motion. It is important to break down the motion into its horizontal and vertical components and consider factors such as initial velocity, acceleration, and time.

4. Can you give an example of a 2D kinematics problem?

One example of a 2D kinematics problem is a ball being thrown at an angle from the ground. The initial velocity of the ball, the angle at which it is thrown, and the acceleration due to gravity can be used to calculate the ball's maximum height, range, and time of flight.

5. How does 2D kinematics differ from 1D kinematics?

2D kinematics differs from 1D kinematics in that it takes into account both horizontal and vertical components of an object's motion, whereas 1D kinematics only considers motion in one direction. This means that 2D kinematics involves vector quantities, such as displacement, velocity, and acceleration, whereas 1D kinematics only involves scalar quantities.

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