Math, asked by ghanshyamc184, 1 month ago

A moving particle starts at an initial position r(0)= (1,0,0) with initial velocity (0)=(-/+k. its acceleration is a(t) = 4tl+ 6tj + k. Determine its velocity and position when t = 2​

Answers

Answered by legendboie81
1

Answer:

iven by

v(t) = h 2t, et

, 3 cost i

(1) Find the position vector r(t) of the particle.

Recall: Let the position of an object be given by r = hf(t), g(t), h(t)i, for t ≥ 0 . The velocity of the

object is v(t) = hf

0

(t), g0

(t), h0

(t)i. If v(t) = hx(t), y(t), z(t)i

r(t) = Z

x(t) dt + C1,

Z

y(t) dt + C2,

Z

z(t) dt + C3

r(t) = Z

2t dt + C1,

Z

e

t

dt + C2,

Z

3 cost dt + C3

=

t

2 + C1, et + C2, 3 sin t + C3

 

Since r(0) = *

0

2 + C1

| {z }

C1

, e0 + C2

| {z }

1+C2

, 3 sin 0 + C3

| {z }

C3

+

= h0, 1, 0i, we have C1 = C2 = C3 = 0.

The position vector r(t) of the particle is

r(t) =  

t

2

, et

, 3 sin t

 

(2) Find the acceleration a(t) of the particle.

Recall: If the velocity of the object is v(t) = hx(t), y(t), z(t)i, then the acceleration of the object is

a(t) = hx

0

(t), y0

(t), z0

(t)i

The acceleration a(t) of the particle is

v(t) = h (2t)

0

, (e

t

)

0

, (3 cost)

0

i = h 2, et

, −3 sin t i

(3) Which of the following points are on the curve of r

( 1, 2, 3 ), ( 4, e2

, 3 sin(2) ), ( 0, 0, 0 )?

If some point is on the curve, find the speed of the particle at that point.

Recall: A point (a, b, c) in on the curve of r = hf(t), g(t), h(t)i if there exists t such that

(a, b, c) = (f(t), g(t), h(t)).

A point (a, b, c) in on the curve of r(t) = ht

2

, et

, 3 sin ti if there exists t such that

(a, b, c) = (t

2

, et

, 3 sin t)

• If (1, 2, 3) = (t

2

, et

, 3 sin t), then 1 = t

2

, 2 = e

t

, and 3 = 3 sin t. The equation

1 = t

2

implies t = 1, but 2 6= e

1 = e. The point (1, 2, 3) is not on the curve of r(t).

• If ( 4, e2

, 3 sin(2) ) = (t

2

, et

, 3 sin t), then 4 = t

2

, e

2 = e

t

, and 3 sin(2) = 3 sin t.

For t = 2 all three equations are satisfied. The point ( 4, e2

, 3 sin(2) ) is on the

curve of r(t).

• If (0, 0, 0) = (t

2

, et

, 3 sin t), then 0 = t

2

, and 0 = e

t

, and 0 = 3 sin t. The equation

0 = t

2

implies t = 0, but 0 6= e

0 = 1. The point (0, 0, 0) is not on the curve of r(t).

the speed of the object

to the curve of r(t)

at time t = 2, i.e.

at the point ( 4, e2

, 3 sin(2) )

= |v(2)| = |h 2 · 2, e2

, 3 cos 2 i|

=

p

4

2 + (e

2

)

2 + (3 cos 2)2 =

16 + e

4 + 9 cos2 2

Step-by-step explanation:

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