Answer-
The rider’s displacement during this time is 17.01 m
Solution-
As we know,
Where s = displacement,
u= initial velocity = 0, as the rider starts from rest
a= acceleration = 0.5 m/s²
t= time taken = 8.4 s
Putting the values,
Mackenzie has a loyalty card good for a 20% discount at her local hardware store. What would her total in dollars and cents be,
2 answers:
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Answer:
(a) -dA/dt = kA², A₀ = 10
(b) A =10/(1+ kt)
(c) t > 60 h
Step-by-step explanation:
(a) Find the IVP
A differential equation with an initial condition y₀ = f(x₀) is called an initial value problem.
The rate of decrease of A is proportional to A², and A₀ = 10, so the IVP is
-dA/dt = kA², A₀ = 10
(b) Solve the IVP
Apply the initial condition: A₀ = 10 (when t = 0)
(c) Find the time when A(t) < 1
(i) Find the value of k (A₁₀ = 4)
(ii) Find t when A < 1
The figure below shows the graph of A vs t.
In the given graph point B is a relative maximum with the coordinates (0, 2).
The given function is .
In the given graph, we need to find which point is a relative maximum.
<h3>What are relative maxima?</h3>The function's graph makes it simple to spot relative maxima. It is the pivotal point in the function's graph. Relative maxima are locations where the function's graph shifts from increasing to decreasing. A point called Relative Maximum is higher than the points to its left and to its right.
In the graph, the maximum point is (0, 2).
Therefore, in the given graph point B is a relative maximum with the coordinates (0, 2).
To learn more about the relative maximum visit:
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