How can you drop two eggs the fewest amount of times, without them breaking?
1 answer:
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Answer:
a) aA = - 13.33 mm/s²
aB = - 20 mm/s²
b) aD = - 13.33 mm/s²
c) vB = 70 mm/s
d) xB = 440 mm
Explanation:
Given
The initial speed of B is: v₀B = 150 mm/s
Distance moved by A is: xA = 240 mm
Velocity of A is: vA = 60 mm/s
Assuming:
Displacement of blocks are denoted by:
A = xA
B = xB
C = xC
D = xD
From the pic shown, the total length of the cable is:
xB + (xB - xA) + 2*(d - xA) = L
⇒ 2*xB - 3*xA = L - 2*d
where L - 2*d is constant. Differentiating the above equation with respect to time:
d(2*xB)/dt - d(3*xA)/dt = 0
⇒ 2*vB - 3*vA = 0 (i)
Substituting in equation (i)
2*(150 mm/s) - 3*vA = 0
⇒ v₀A = 100 mm/s (initial speed of A)
Then, we use the equation
vA² = v₀A² + 2*aA*xA
Substituting the values in above equation:
(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)
⇒ aA = - 13.33 mm/s²
If 2*vB - 3*vA = 0
Differentiating the above equation with respect to time:
d(2*vB)/dt - d(3*vA)/dt = 0
⇒ 2*aB - 3*aA = 0 (ii)
Substituting in equation (ii)
2*aB - 3*(- 13.33 mm/s²) = 0
⇒ aB = - 20 mm/s²
b) From the pic shown,
xD - xA = constant
If we apply
d(xD)/dt - d(xA)/dt = 0
⇒ vD - vA = 0
then
d(vD)/dt - d(vA)/dt = 0
⇒ aD - aA = 0
⇒ aD = aA = - 13.33 mm/s²
c) We use the formula
vB = v₀B + aB*t
Substituting the values in above equation:
vB = 150 mm/s + (- 20 mm/s²)*(4 s)
⇒ vB = 70 mm/s
d) We apply the equation
xB = v₀B*t + 0.5*aB*t²
Substituting the values in above equation:
xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²
⇒ xB = 440 mm
