Answer:
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Explanation:
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
Answer:
Explanation:
We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .
rate of volume flow of one stream
= cross sectional area x velocity
= 8.4 x 3.5 x 2.2 = 64.68 m³ /s
rate of volume flow of other stream
= 6.6 x 3.6 x 2.7
= 64.15 m³ /s
rate of volume flow of rive , if d be its depth
= 11.2 x d x 2.8
= 31.36 d
volume flow of river = Total of volume flow rate of two streams
31.36 d = 64.15 + 64.68
31.36 d = 128.83
d = 4.10 m /s .
