# Can somebody help translate this question? 2D Kinematics.

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?

(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

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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PeterO
Homework Helper
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?

(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

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.

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
Homework Helper
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
Mentor
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).

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. :)

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?

(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

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
Staff Emeritus