An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far wi
1 answer:
Answer:
218m
Explanation:
Given parameters:
Initial velocity = 3.3m/s
acceleration = 3.7m/s²
time = 10s
Unknown:
How far will it travel during the time of acceleration = ?
Solution:
We use of the kinematics equations to solve this problem;
S = ut + at²
S is the distance
u is the initial velocity
t is the time
a is the acceleration
So;
S = (3.3x10) + ( x 3.7 x 10²) = 218m
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The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.![\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10-0.8v%20%5C%5C%5C%5C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%3Ddt%20%5C%5C%5C%5C%20%5Cint_%7B16.5%7D%5E%7Bv%7D%20%5C%2C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%20%20%3D%20%5Cint_%7B0%7D%5E%7Bt%7D%20dt%20%5C%5C%5C%5C%20-%20%5Cfrac%7B1%7D%7B0.8%7D%20%5Bln%2810-0.8v%29%5D_%7B16.5%7D%5E%7Bv%7D%3Dt%20%5C%5C%5C%5C%20ln%20%5Cfrac%7B10-0.8v%7D%7B-3.2%7D%3D-0.8t%20%5C%5C%5C%5C%20%20%5Cfrac%7B0.8v%20-10%7D%7B3.2%7D%20%20%3De%5E%7B-0.8t%7D%20%5C%5C%5C%5C%200.8v%20%3D%2010%20%2B%203.2e%5E%7B-0.8t%7D%20%5C%5C%5C%5C%20v%3D12.5%2B4e%5E%7B-0.08t%7D)
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
![d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft](https://tex.z-dn.net/?f=d%3D%5Cint_%7B0%7D%5E%7B5.7%7D%20%5C%2C%20%2812.5%2B4e%5E%7B-0.8t%7D%29dt%20%5C%5C%5C%5C%20%3D%2012.5%285.7%29%2B%20%5Cfrac%7B4%7D%7B%28-0.8%29%7D%5Be%5E%7B-0.8t%7D%5D_%7B0%7D%5E%7B5.7%7D%20%5C%5C%5C%5C%20%3D71.25-5%280.0105-1%29%20%3D76.198%20%5C%2C%20ft)
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.
The velocity function is
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
The depth of the lake is
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
Answer:
greater than your true weight
Explanation:
When going up in an elevator the acceleration of the elevator is added to the acceleration due to gravity. This will increase the reading on the scale.
The expression of the resultant weight will be
where,
m = Mass of the person
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the elevator.
Hence, the reading on the scale is <u>greater than your true weight.</u>