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前女友ample Questions
前女友ample Question #1 :Velocity
If during the first 15 seconds of its flight the displacement of a spacecraft is given by the equation 21.25x+ x2feet, what is its velocity at 10 seconds, given that its initial velocity is 0?
None of the other answers
1.25 * 212.5+ 20 feet/second
1.25 * 212.5* log(2) + 100 feet/second
21.25+ 20 feet/second
1.25 * 212.5* log(2) + 20 feet/second
1.25 * 212.5* log(2) + 20 feet/second
To find the velocity function based on displacement, use the first derivative of f(x) = 21.25x+ x2. f'(x) = 1.25 * ln(2) * 21.25x+ 2x
The velocity at x = 10 would therefore be: f'(10) = 1.25 * log(2) * 21.25 * 10+ 2 * 10 = 1.25 * 212.5* log(2) + 20 feet/second
前女友ample Question #2 :Velocity
What is the instantaneous velocity at time t = π/2 of a particle whose positional equation is represented by s(t) = 12tan(t/2 + π)?
6
6√2
12
3√2
12√2
12
The instantaneous velocity is represented by the first derivative of the positional equation.
v(t) = s'(t) = 12 * (1/2) sec2(t/2 + π) = 6sec2(t/2 + π) = 6/cos2(t/2 + π) = 6/((–1)2cos2(t/2)) = 6/cos2(t/2)
Based on the nature of the cosine, we know that 6/cos2(t/2 + π) = 6/((–1)2cos2(t/2)) = 6/cos2(t/2)
To solve, v(π/2) = 6/cos2(t/2) = 6/((1/√2)2) = 6/(1/2) = 12
前女友ample Question #3 :Velocity
什么是瞬时速度时间t =π/ 2 of a particle whose positional equation is represented by s(t) = 12cos2(t/2 + π)?
–12
2√3
0
2
–6
–6
The instantaneous velocity is represented by the first derivative of the positional equation. This is found by using the chain rule both on the square of the cosine function and the function itself.
v(t) = s'(t) = 12 * 2 cos(t/2 + π) * (–sin(t/2 + π)) * (1/2) = –12cos(t/2 + π)sin(t/2 + π)
Given what we know about the cosine and sine functions, we know cos(t/2 + π) = –cos(t/2) and sin(t/2 + π) = –sin(t/2)
Therefore, v(t) = –12(–sin(t/2))(–cos(t/2)) = –12sin(t/2)cos(t/2)
v(π/2) = –12sin(π/4)cos(π/4) = –12(1/√2)(1/√2) = –12 * (1/2) = –6
前女友ample Question #4 :Velocity
A weight hanging from a spring is stretched down 3 units beyond its rest position and released at timet=0 to bob up and down. Its position at any later timetis
What is its velocity at time?
前女友ample Question #5 :Velocity
The positionof a particle at timeis given by. What is the particle's velocity at time
The velocity function is given by the derivative of the position function. So here. Plugging 3 in forgives 16.
前女友ample Question #6 :Velocity
The position of a particle is given by. Find the velocity at.
The velocity is given as the derivative of the position function, or
.
We can use the quotient rule to find the derivative of the position function and then evaluate that at. The quotient rule states that
.
In this case,and.
We can now substitute these values in to get
.
Evalusting this atgives us.
So the answer is.
前女友ample Question #1 :How To Find Velocity
Find the velocity function if the position function is given as:.
有三个在这probl条款em that has to be derived. The derivative of the position function, or the velocity function, represents the slope of the position function.
The derivative ofcan be solved by using the power rule, which is:
Therefore. the derivative ofis.
The derivative ofisby using the constant multiple rule.
The derivative ofis由于衍生品of constants equal to zero.
前女友ample Question #8 :Velocity
Find the velocity function given the position function:.
The derivatives can be solved term by term.
First, find the derivative of. This can be done by the power rule.
Find the derivative of. Rewrite this asto compute by power rule.
Therefore, the velocity function is:
前女友ample Question #9 :Velocity
Find the velocity of a function if the acceleration is:.
To find the velocity given the acceleration function, we will need to integrate the acceleration function.
前女友ample Question #10 :Velocity
Find the velocity atif the acceleration function is:.
The velocity function can be obtained by integrating the acceleration function.
Since we are finding the velocity at, substitute this into the velocity function.
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