a) Function: [tex]f(t)=940-7.5t[/tex]
b) Independent variable: time, domain: [tex]0\leq t \leq 125.3[/tex]
c) Dependent variable: distance from the bus stop, range: [tex]0\leq f(t) \leq 940[/tex]
d)
[tex]f(0)=940 ft[/tex] represents Bill's distance from the bus stop at t = 0 (initial instant)
[tex]f(60.25)=488ft[/tex] represents Billy's distance from the bus stop at t = 60.25 s
e)
i) After 23.6 seconds, Billy's distance from the bus stop is 763 feet.
ii) Change in distance: -172.5 ft
f)
Billy's distance from the bus stop is 150 ft after t = 105.3 s.
Step-by-step explanation:
a)
Here we want to write a function that represents Billy's distance from the bus stop at time t.
We know that at the beginning, Billy's distance from the bus stop is
d = 940 feet
And then, he walks with a speed of
v = 7.5 ft/s
towards the bus stop.
Therefore, the distance from the bus stop can be represented with the following function:
[tex]f(t)=d-vt=940-7.5t[/tex]
b)
The independent quantity of a function is the variable whose value does not depend on the other: that is, in this case, it is the time (t).
The domain of a function is the set of possible values that the independent variable can take.
In this problem, t is restricted between 0 (minimum value) and t', the time at which Billy reaches the bus stop, which is when
f(t')=0
Which can be found as follows:
[tex]940-7.5t'=0\\940=7.5t'\\t'=\frac{940}{7.5}=125.3 s[/tex]
So the domain is [tex]0\leq t \leq 125.3 s[/tex]
c)
The dependent variable of a function is the variable whose value depends on the other one: in this case, the dependent variable is the distance from the bus stop, f(t).
The range of a function is the set of possible values that the dependent variable can have.
In this case, the distance from the bus stop has a minimum value of 0 (final moment) and a maximum value of 940 (initial moment): therefore, the range of the function is
[tex]0\leq f(t) \leq 940 ft[/tex]
d)
Billy's distance from the bus stop is
[tex]f(t)=940-7.5t[/tex]
where
t is the time (measured in seconds)
Therefore:
[tex]f(0)=940[/tex] represents Bill's distance from the bus stop at t = 0 (initial instant)
[tex]f(60.25)=940-7.5\cdot 60.25=488 ft[/tex] represents Billy's distance from the bus stop at t = 60.25 s
e)
i) Here we want to write Billy's distance from the bus stop after he has walked 23.6 seconds.
The initial distance of Billy from the bus stop is
d = 940 feet
Therefore, Billy's distance from the bus stop at time t can be represented with the following function:
[tex]f(t)=d-vt[/tex]
where
v = 7.5 ft/s is Billy's speed
Substituting,
[tex]f(t)=940-7.5t[/tex]
And substituting t = 23.6 s, we find:
[tex]f(23.6)=940-(7.5)(23.6)=763 ft[/tex]
So, after 23.6 seconds, Billy's distance from the bus stop is 763 feet.
ii)
For this part, we have to find first Billy's distance at t = 25 s and at t = 48 s, and calculate the difference between the two values.
For t = 25 s, Billy's distance from the bus stop is:
[tex]f(25)=940-7.5\cdot 25=752.5 ft[/tex]
At t = 48 s, Billy's distance from the bus stop is
[tex]f(48)=940-7.5\cdot 48 =580 ft[/tex]
Therefore, the change in distance between the two instants is:
[tex]\Delta f = f(48)-f(25)=580-752.5=-172.5 ft[/tex]
f)
In this case, we want to solve the equation
[tex]f(t)=150[/tex]
Where 150 represents the distance of Billy from the bus stop at a certain time t. Therefore, here we want to find the time t at which Billy's distance from the bus stop is 150 ft.
Substituting and solving, we find:
[tex]940-7.5t=150\\940-150=7.5t\\790=7.5t\\t=\frac{790}{7.5}=105.3 s[/tex]
Therefore, Billy's distance from the bus stop is 150 ft after t = 105.3 s.
Learn more about functions:
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