Question

The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is measured in seconds. What is the maximum velocity of the object in the time interval [0, 4]?

Answer 1

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 $s(t) = -t^3+3t^2+7t +4$, 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,

$s(t) = -t^3+3t^2+7t +4$

Derivate w.r.t. time,

$v(t)=s'(t) = -3t^2+6t+7$

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

$x=-\frac{b}{2a}$

Where, a=-3, b=6 and c=7

$t=-\frac{6}{2(-3)}$

$t=-\frac{6}{-6}$

$t=1$

As 1 lie between interval [0,4]

Substitute t=1 in the function,

$v(t)= -3(1)^2+6(1)+7$

$v(t)= -3+6+7$

$v(1)=10$

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 2

Answer:

$v_{max} = 10\,\frac{ft}{s}$ at $t = 1\,s$.

Step-by-step explanation:

The velocity function is found by deriving once:

$v(t) = -3\cdot t^{2}+6\cdot t +7$

The acceleration function is determined by deriving again:

$a(t) = -6\cdot t + 6$

The critical point of the velocity function is computed by equalizing the acceleration function to zero and clearing t:

$-6\cdot t + 6 = 0$

$t = 1\,s$

The Second Derivative Test is done by deriving the acceleration function and checking the critical point hereafter:

$\dot a (t) = -6$

Which indicates that critical point leads to maximum velocity, which is:

$v(1\,s) = -3\cdot (1\,s)^{2} + 6\cdot (1\,s) + 7$

$v_{max} = 10\,\frac{ft}{s}$