3 years ago

Help please ASAP???

Mathematics

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

B. y = -2x

and

C. y = 5x

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2 years ago

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 meas

saw5 [17]

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.

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3 years ago

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2 years ago

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Answer:

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3 years ago

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