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Arte-miy333 [17]
2 years ago
6

The equation r(t)= (2t)i + (2t-16t^2)j is the position of a particle in space at time t. Find the angle between the velocity and

acceleration vectors at time t=0.

Mathematics
1 answer:
ankoles [38]2 years ago
4 0

Answer:

The answer is 135 degrees.

Step-by-step explanation:

As we are given the position. If we take the <u>derivative</u>, we get the velocity vector. If we take the <u>derivative</u> again, we find the acceleration vector of the particle.

r(t)=(2t)i+(2t-16t^{2})j

V(t)=2i+(2-32t)j

a(t)=-32j

At time t=0;

v(t)=2i+2j

a(t)=-32j

As i attach in the picture the angle between the velocity and acceleration vector is (45+90)=135 degrees

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R = (- 3.5, - 7 )

Step-by-step explanation:

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The average sea level in 1900 at London bridge was 33 feet. In 1990 it was 33.08 feet. Use linear interpolation or extrapolation
eduard

Answer:

In 1956, the height will be 33.05 ft

Step-by-step explanation:

Let +x+ = number of years after 1900

Calll 1900 +x+=+0+

Then 1990 is +x+=+90+

Let +y+ = height in feet

--------------------------

You are gven the points

( 0, 33 )

( 90, 33.08 )

Use general point-slope formula

+%28+y+-+33+%29+%2F+%28+x+-+0+%29+=+%28+33.08+-+33+%29+%2F+%28+90+-+0+%29+

Multiply both sides by +90x+

+90%2A%28+y+-+33+%29+=+.08x+

+90y+-+2970+=+.08x+

+90y+=+.08x+%2B+2970+

+y+=+.000889x+%2B+33+ ( equation to use )

--------------------------

Check:

does it go through ( 90, 33.08 ) ?

+33.08+=+.000889%2A90+%2B+33+

+33.08+=+.08001+%2B+33+

+33.08+=+33.08001+

close enough

-------------------------------

(a)

+x+=+61%0D%0A%7B%7B%7B+y+=+.000889x+%2B+33+

+y+=+.000889%2A61+%2B+33+

+y+=+.0542+%2B+33+

+y+=+33.0542+

In 1961, the height will be 33.05 ft

8 0
2 years ago
In exercises 1 and 2, find m&lt;1 and m&lt;2. Tell which theorem you used in each case.
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Answer:

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From the given picture,

lines 'm' and 'n' are the parallel lines and line 't' is the transverse line.

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