A particle leaves its orgin with a velocity of 4 i m/s and a constant acceleration of (-2i + .5j) m/s^2. At the time the particle reaches its maximum X corrdinate a) what is its velocity? b) what is its position vector?

I'm not sure which equations i should use. I am confused by this problem

Explain Bot Explain Bot answered
9 months ago

To solve this problem, we will need to use the equations of motion in two dimensions. These equations relate the position, velocity, and acceleration of an object.

The equations of motion in two dimensions are:

1. Position equation: r = r0 + v0t + (1/2)at^2
2. Velocity equation: v = v0 + at
3. Acceleration equation: a = constant

Given that the particle leaves its origin with a velocity of 4i m/s and a constant acceleration of (-2i + 0.5j) m/s^2, we can plug these values into the equations of motion.

a) To find the particle's velocity when it reaches its maximum X coordinate, we need to find the time it takes for the particle to reach this position. The maximum X coordinate implies that the particle has zero velocity in the X direction.

Using the velocity equation, we can set the X component of the particle's velocity equal to zero:

0 = 4i + (-2i + 0.5j)t

Simplifying the equation, we get:

0 = 2i + 0.5j)t

Solving for t, we get:

t = -4i/j = -4/0.5 = -8 seconds

Now that we have the time, we can find the particle's velocity at that time:

v = v0 + at
= 4i + (-2i + 0.5j)(-8)
= 4i + 16i - 4j
= 20i - 4j m/s

Therefore, the particle's velocity when it reaches its maximum X coordinate is 20i - 4j m/s.

b) To find the position vector at the time the particle reaches its maximum X coordinate, we can use the position equation:

r = r0 + v0t + (1/2)at^2

Since the particle starts at the origin (r0 = 0), we can simplify the equation:

r = v0t + (1/2)at^2
= (4i)(-8) + (1/2)(-2i + 0.5j)(-8)^2
= -32i + (-4i + j)(64)
= -32i + (-256i + 64j)
= -288i + 64j m

Therefore, the particle's position vector at the time it reaches its maximum X coordinate is -288i + 64j m.