A watershed is divided into two sub-basins A and B. The surface runoff from sub-basin A enters a channel at point 1 and flows to
1 answer:
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(a) V(A/B) = (14 i - 17.32 J) m/s
(b) acc(A/B) = ( 5.11 i + 5.13 j ) m/s²
<u>Explanation:</u>
We will solve with respect to Cartesian vector form.
So,
V(A)= (24i) m/s
acc(A) = (4i) m/s²
There are two components of Car B, cos 60⁰ and sin 60⁰
V(B) = 20 cos 60° i + 20 sin 60° j
V(B) = (10 i + 17.32 j ) m/s
The car B moves along a curve, so it will have a tangential acceleration and a normal acceleration.
The tangential acceleration, a(t) = 5 m/s²
Normal acceleration, a(n) =
So,
For the tangential acceleration, the acceleration is slowing down. So,
a(t) = (-5 cos 60° i - 5 sin 60° j ) m/s²
a(t) = ( -2.5 i - 4.33 j) m/s²
For normal acceleration, it towards center. So,
a(n) = (1.6 sin 60° i - 1.6 cos 60° j) m/s²
a(n) = (1.39 i - 0.8 j ) m/s²
Total acceleration of Car B:
acc(B) = a(t) + a(n)
acc(B) = ( -2.5 i - 4.33 j) m/s² + (1.39 i - 0.8 j ) m/s²
acc(B) = (-1.11i - 5.13 j ) m/s²
(a) V(A/B) = ?
V(A) = V(B) + V(A/B)
(24i) m/s = (10 i + 17.32 j ) m/s + V(A/B)
V(A/B) = (14 i - 17.32 J) m/s
(b) acc(A/B) = ?
acc(A) = acc(B) + acc(A/B)
(4i) m/s² = (-1.11i - 5.13 j ) m/s² + acc(A/B)
acc(A/B) = ( 5.11 i + 5.13 j ) m/s²