Can somebody help translate this question? 2D Kinematics.

[DavidAp]

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.

 

[PeterO]

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.

 

[DavidAp]

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?

 

[PeterO]

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.

 

[gneill]

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).

 

[DavidAp]

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. :)

 

[theunloved]

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 ?

 

[SammyS]

The direction of the velocity is the direction of travel.

 

[theunloved]

The direction of the velocity is the direction of travel.

Thank you !