Collar P slides outward at a constant relative speed along rod AB, which rotates counterclockwise with a constant angular veloci
1 answer:
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
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Answer:
a)
Explanation:
- Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:
- Since the car starts from rest, v₀ =0.
- We know the value of t = 5 sec., but we need to find the value of a.
- Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:
- Replacing a and t in (1):
- Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
- Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:
- Replacing v₀, at and t in (1), we have:
- Therefore, as the truck travels twice as far as the car, the right answer is a).