A particle has ⃗r(0) = (4 m)j and ⃗v(0) = (2 m/s)i.
If its acceleration is constant and given by ⃗a = −(2 m/s2) (i +j), at what time t does the particle first cross the x axis?
Answer in units of s.
part 2. At what time t is the particle moving parallel to the y axis; that is, in the ˆ direction?
Answer in units of s.
part 2. At what time t is the particle moving parallel to the y axis; that is, in the j direction?
Answer in units of s.
im really lost and if anyone could atleast tell me which equation to use it would be greatly appreciated
To find the time at which the particle first crosses the x-axis (part 1), we need to determine when the y-coordinate of the particle becomes zero.
We know that the particle starts at ⃗r(0) = (4 m)j, which means the initial y-coordinate is 4 m. The acceleration is given as ⃗a = −(2 m/s²)(i + j). This means that both the x-component and y-component of the acceleration are constant and equal to -2 m/s².
To find the time at which the particle crosses the x-axis, we can use the equations of motion. The y-component of the position can be determined using the equation:
y(t) = y(0) + v(0)y * t + (1/2) * a * t²
where y(t) is the y-coordinate of the particle at time t, y(0) = 4 m is the initial y-coordinate, v(0)y = 0 m/s is the initial y-component of velocity, a = -2 m/s² is the constant acceleration, and t is the unknown time.
Since the particle is crossing the x-axis, the y-coordinate (y(t)) will be zero. Plugging in the known values, we get:
0 = 4 m + 0 m/s * t + (1/2)(-2 m/s²) * t²
Simplifying the equation, we have:
0 = 4 - t - t²
Rearranging the terms, we get a quadratic equation:
t² + t - 4 = 0
We can solve this equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
where a = 1, b = 1, and c = -4. Plugging in these values, we get:
t = (-1 ± √(1² - 4(1)(-4))) / (2 * 1)
Simplifying further, we have:
t = (-1 ± √(1 + 16)) / 2
t = (-1 ± √17) / 2
Since we are looking for the positive time value when the particle first crosses the x-axis, we take the positive square root:
t = (-1 + √17) / 2
Thus, the time at which the particle first crosses the x-axis is approximately -0.56 s.
To find the time at which the particle is moving parallel to the y-axis (part 2), we need to determine when the x-component of the particle's velocity becomes zero.
We know that the particle starts with ⃗v(0) = (2 m/s)i, which means the initial x-component of velocity is 2 m/s. The acceleration remains constant as ⃗a = −(2 m/s²) (i + j).
The x-component of velocity (v(x)) can be determined using the relation:
v(x) = vₓ(0) + aₓ * t
where v(x) is the x-component of velocity, vₓ(0) = 2 m/s is the initial x-component of velocity, aₓ = -2 m/s² is the constant x-component of acceleration, and t is the unknown time.
To find the time when v(x) becomes zero, we set v(x) equal to zero and solve for t:
0 = 2 m/s + (-2 m/s²) * t
Simplifying the equation, we have:
0 = 2 - 2t
Rearranging the terms, we get:
2t = 2
t = 1 s
Thus, the time at which the particle is moving parallel to the y-axis is 1 second.