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aleksandr82 [10.1K]
3 years ago
5

If u = <-7, 6> and v = <-4, 17>, which vector can be added to u + 3v to get the unit vector <1, 0> as the resu

ltant vector?
Mathematics
1 answer:
ivann1987 [24]3 years ago
7 0

With u = <-7, 6> and v = <-4, 17>, we have

u + 3v = <-7, 6> + 3 <-4, 17> = <-7, 6> + <-12, 51> = <-19, 57>

We want to find a vector w such that

u + 3v + w = <1, 0>

Subtract u + 3v from both sides to get

w = <1, 0> - (u + 3v) = <1, 0> - <-19, 57>

w = <20, -57>

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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
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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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