Two Dimensional Kinematics probelms

[riken9]

Homework Statement

1. At t = 0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (2.0i - 4.0j) m/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle?

2. A particle starts from the origin at t = 0 with a velocity of 6.0 m/s and moves in the xy plane with a constant acceleration of (-2.0 + 4.0 ) m/s2. At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?

Homework Equations

rf = ri + vt + 1/2at^2
vf = vi + at

The Attempt at a Solution

1. rf = ri + vt + 1/2at^2
t = 2.25 [seconds]

vf = (9j) + [2i - 4j)t

Sub t into there and i get 4 m/s But the answer is: 10 m/s


2. v(initial)=6i m/s
v(final)=?
a= (-2i + 4j) m/s^2
starts at coordinate (0,0)
need to find max x point

rf = 6i + [-1i + 2j]t^2

vf = vi + at
How do i find t if i don't have final velocity? Answer is: 20m.

 

[TSny]

Hello, riken9. Welcome to PF!

For the first problem, your answer for the time is incorrect. Can you show your work so we can find the mistake?

For the second problem, there is not enough information given. Did they give you the direction of the velocity at t = 0? [EDIT: OK, it looks like from your work that the initial velocity is in the x direction. I think you will find things will be easier if you write out separate equations for the x and y components of the motion.]

 

[riken9]

for the first question rf = ri + vt + 1/2at^2, i did 15 = 0 + 9t - 2t^2 and got 2.25 seconds. I also tried 15 = 9j + [i - 2j]t^2 and got 1.27 seconds, which is also wrong.

And for the second problem i tried that still couldn't get it right.

 

[TSny]

for the first question rf = ri + vt + 1/2at^2, i did 15 = 0 + 9t - 2t^2

In the above equation you have mixed together x quantities (blue) and y quantities (red). Instead, write out equations for just the x component of motion and separate equations for the y component of motion. For example

x = x_i + v_xit + (1/2)a_xt²

y = y_i + v_yit + (1/2)a_yt²

or for velocity,

v_x = v_xi + a_xt

v_y = v_yi + a_yt

So, try using x = x_i + v_xit + (1/2)a_xt² to find the time for the first problem.

 

[Nickie]

I would like to clarify some key concepts in solving these two dimensional kinematics problems. First, it is important to understand that the given acceleration vector (2.0i - 4.0j) m/s2 in problem 1 means that the particle is accelerating in the negative x direction and downward in the y direction. Similarly, in problem 2, the acceleration vector (-2.0i + 4.0j) m/s2 indicates acceleration in the negative x direction and upward in the y direction.

In order to solve problem 1, we can use the equations of motion to find the time it takes for the particle to reach the x coordinate of 15 m. Once we have the time, we can then use the velocity equation to find the final velocity of the particle. It is important to carefully consider the direction of the acceleration and velocity vectors in order to arrive at the correct answer.

For problem 2, we can use the same approach but instead of finding the final velocity, we need to find the maximum x coordinate of the particle. This can be done by setting the x component of the velocity equation to 0 and solving for time. Once we have the time, we can then use the position equation to find the x coordinate at that time.

In both problems, it is important to pay attention to units and to carefully substitute and manipulate equations in order to arrive at the correct answer. As a scientist, it is crucial to understand and apply the fundamental principles of kinematics in order to solve these types of problems accurately.