Respuesta :
Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by [tex]s(t) = -t^3+3t^2+7t +4[/tex], where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
[tex]s(t) = -t^3+3t^2+7t +4[/tex]
Derivate w.r.t. time,
[tex]v(t)=s'(t) = -3t^2+6t+7[/tex]
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
[tex]x=-\frac{b}{2a}[/tex]
Where, a=-3, b=6 and c=7
[tex]t=-\frac{6}{2(-3)}[/tex]
[tex]t=-\frac{6}{-6}[/tex]
[tex]t=1[/tex]
As 1 lie between interval [0,4]
Substitute t=1 in the function,
[tex]v(t)= -3(1)^2+6(1)+7[/tex]
[tex]v(t)= -3+6+7[/tex]
[tex]v(1)=10[/tex]
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Answer:
[tex]v_{max} = 10\,\frac{ft}{s}[/tex] at [tex]t = 1\,s[/tex].
Step-by-step explanation:
The velocity function is found by deriving once:
[tex]v(t) = -3\cdot t^{2}+6\cdot t +7[/tex]
The acceleration function is determined by deriving again:
[tex]a(t) = -6\cdot t + 6[/tex]
The critical point of the velocity function is computed by equalizing the acceleration function to zero and clearing t:
[tex]-6\cdot t + 6 = 0[/tex]
[tex]t = 1\,s[/tex]
The Second Derivative Test is done by deriving the acceleration function and checking the critical point hereafter:
[tex]\dot a (t) = -6[/tex]
Which indicates that critical point leads to maximum velocity, which is:
[tex]v(1\,s) = -3\cdot (1\,s)^{2} + 6\cdot (1\,s) + 7[/tex]
[tex]v_{max} = 10\,\frac{ft}{s}[/tex]
