Answer:
Δx = 100 − 50/a
a = 2/3
Explanation:
Since both particles undergo constant acceleration, their positions and velocities can be modeled using kinematics. Two such kinematic equations are:
x = x₀ + v₀ t + ½ at²
v = v₀ + at
where x is the final position, x₀ is the initial position, v₀ is the initial velocity, v is the final velocity, a is acceleration, and t is time.
Particle A moves at steady speed, so its acceleration is zero. If we say that particle A starts at the origin, then the position of particle A at time t is:
x₁ = 0 m + (10 m/s) t + ½ (0 m/s²) t²
x₁ = 10t
And the velocity of particle A at time t is:
v₁ = 10 m/s + (0 m/s²) t
v₁ = 10
Particle B starts at rest, so v₀ is zero. Since particle B starts moving 100 meters ahead of particle A, then the position of particle B at time t is:
x₂ = 100 m + (0 m/s) t + ½ at²
x₂ = 100 + ½ at²
And the velocity of particle B at time t is:
v₂ = 0 m/s + at
v₂ = at
The difference between the positions of the particles is:
Δx = x₂ − x₁
Δx = (100 + ½ at²) − (10t)
Δx = ½ at² − 10t + 100
This is a quadratic equation. The vertex of a parabola occurs at x = -b / 2a, where a is the coefficient of the squared term and b is the coefficient of the linear term. So the minimum distance between the particles occurs at time:
t = -(-10) / 2 (½ a)
t = 10 / a
The velocity of particle B at this time is:
v₂ = a (10/a)
v₂ = 10
Therefore, the distance between the particles is a minimum when they have the same speed.
This minimum distance is:
Δx = 100 + ½ a (10/a)² − 10 (10/a)
Δx = 100 + 50/a − 100/a
Δx = 100 − 50/a
If the shortest distance is 25 m, then the value of a is:
25 = 100 − 50/a
50/a = 75
a = 2/3
