Acceleration is (-2.0i + 4.0j), velocity is 12 m/s

In summary: V_{ix}[/tex] = 12 m/s[itex]V_{iy}[/tex] = 0 m/sNow, using the equation [itex]rf=ri + vit + 1/2at^2[/itex], we can find the x and y coordinates of the particle at any given time. However, we are interested in finding the x coordinate at the instant the y coordinate is 18 m. So, let's set [itex]y_f = 18 m[/itex] and solve for [itex]x_f[/itex]:18m = 0m + (12 m/s)(t) + 1/2 (-2
  • #1
new^2^physics
13
0

Homework Statement



At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction and moves in the xy plane with a constant acceleration of (-2.0i + 4.0j)m/s^2. At the instant the y coordinate of the particle is 18 m, what is the x coordinate of the particle.

Homework Equations



rf=ri + vit + 1/2at2
=(12m/s)(t) + 1/2 (-2.0i + 4.0j)t2

So stuck. Any direction or help would be greatly appreciated!
 
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  • #2
so you know initial velocity to be 12.0i..
now you also know a = dv/dt
and v = dr/dt (r being position)..

Which mathematical (or calculus) tool can you use to get your answer now? and how?
 
  • #3
Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt

dt = 12i / -6.0i + 4.0j
dt = -6.0i - 4.0 j

v = dr/dt
12 i = xi - 18j / -6.0i - 4.0j
12i(-6.0i - 4.0 j) = xi - 18j

oh dear, now I am stuck again and I am not even sure I have done this part correctly.
 
  • #4
new^2^physics said:
Okay, so I am going to play with these equations and try to get closer...

a = dv/dt
(-2.0i + 4.0j)= 12i / dt


that isn't correct...

a=dv/dt
(-2.0i+4.0j)dt=dv
integrating

-(2t)i + (4t)j = v-12

so you get v as a function of time... now proceed in a similar manner...
 
  • #5
new^2^physics said:
At T=0 a particle leaves the origin with a velocity of 12 m/s in the positive x direction...

So, what are [itex]V_{ix}[/tex] and [itex]V_{iy}[/tex]?
 

Related to Acceleration is (-2.0i + 4.0j), velocity is 12 m/s

1. What is the magnitude of acceleration in this scenario?

The magnitude of acceleration is given by the formula |a| = √(ax2 + ay2). Plugging in the values of -2.0 for ax and 4.0 for ay, we get a magnitude of |a| = √(4 + 16) = √20 m/s2.

2. What is the direction of acceleration in this scenario?

The direction of acceleration can be represented by the angle θ with respect to the positive x-axis. Using trigonometric identities, we can calculate θ = tan-1(ay/ax) = tan-1(4/-2) = -63.4° with respect to the positive x-axis.

3. How does the velocity change with respect to time?

Using the formula v = u + at, where u is the initial velocity, t is the time, and a is the acceleration, we can calculate the change in velocity over time. In this scenario, the initial velocity u is 12 m/s and the acceleration a is (-2.0i + 4.0j) m/s2. Therefore, the velocity changes over time according to the equation v = (12 m/s) + (-2.0 m/s2)(t)i + (4.0 m/s2)(t)j.

4. Is the acceleration constant in this scenario?

No, the acceleration is not constant in this scenario. The values of ax and ay are changing, indicating that the acceleration is not constant. However, the magnitude of acceleration (as calculated in question 1) remains constant.

5. How does this scenario relate to Newton's Second Law of Motion?

According to Newton's Second Law of Motion, the force acting on an object is equal to its mass multiplied by its acceleration (F = ma). In this scenario, the acceleration is (-2.0i + 4.0j) m/s2, and the mass is not given. Therefore, we cannot apply Newton's Second Law without knowing the mass of the object.

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