In the figure, particle A moves along the line y = 26 m with a constant velocity of magnitude 2.6 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration of magnitude 0.31 m/s2. What angle θ between and the positive direction of the y axis would result in a collision?

Answer 1

As particle A is moving horizontally along the x-axis, let's find the time t when the x-coordinates of both particles A and B are the same.

Let xA be the x-coordinate of particle A, and xB be the x-coordinate of particle B.

For particle A, we have:

xA = vA * t, where vA = 2.6 m/s is its constant velocity.

For particle B, we have:

xB = vB * cos(θ) * t, where vB = aB * t is its speed, and aB = 0.31 m/s^2 is its constant acceleration.

So, when xA = xB, that's when the two particles collide:

2.6 * t = (0.31 * t) * cos(θ) * t

Thus, we need to find the angle θ that satisfies this equation.

Divide both sides by 2.6 * t:

1 = (0.31 / 2.6) * cos(θ) * t

Now, divide both sides by t:

1 / t = (0.31 / 2.6) * cos(θ)

We see that cos(θ) = (2.6 / 0.31) * (1 / t):

cos(θ) = 8.387 * (1 / t)

Now we can substitute the expression:

cos(θ) = 8.387 * (1 / t)

θ = arccos(8.387 * (1 / t))

Now we just need to find the appropriate value of t.

Since particle B has a constant acceleration aB, and its initial speed is zero, the time it takes to reach a speed of 2.6 m/s (which is the speed of particle A) is:

t = vA / aB = 2.6 / 0.31 ≈ 8.387 s

Finally, we can find the angle θ:

θ = arccos(8.387 * (1 / 8.387)) = arccos(1)

θ = 0

So the angle θ between the y-axis and the direction of particle B would be 0 degrees. Particle B would move straight up along the y-axis in order to collide with particle A.