Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s2. Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, use kinematics to determine:

a. The acceleration of A and B

b. The initial velocities of A and C

c. The change in positions of slider C after 3 s.

a) aB = 240 mm/s² (↓)

aA = - 345 mm/s²

b) For the Block C

v₀ = 130 mm/s

For the Block A

v₀ = - 43.33 mm/s

c) Δx = 727.5 mm

Explanation:

a) For the Block B

v = v₀ + a*t

480 = 0 + a * (2)

aB = 240 mm/s² (↓)

Then we have

3*LA + 4*LB + LC = L

If we apply

d (3*LA + 4*LB + LC) / dt = dL/dt

3*vA + 4*vB + vC = 0

d (3*vA + 4*vB + vC) / dt = d (0) / dt

3*aA + 4*aB + aC = 0

aA = - (4*aB + aC) / 3

aA = - (4*240 mm/s² + 75 mm/s²) / 3

aA = - 345 mm/s²

b) For the Block C

v = v₀ + a*t

v₀ = v - a*t

v₀ = 280 - (75) (2) = 130 mm/s

For the Block A

When t = 2 s

vB = 480 mm/s

vC = 280 mm/s

we use the formula

3*vA + 4*vB + vC = 0

3*vA + 4 * (480) + 280 = 0

vA = - 733.33 mm/s

Now, we can apply

v = v₀ + a*t

- 733.33 mm/s = v₀ + ( - 345 mm/s2) (2 s)

v₀ = - 43.33 mm/s

c) We can use the equation

Δx = v₀*t + (1/2) * a*t²

Δx = 130 * (3) + (1/2) (75) (3) ²

Δx = 727.5 mm