The eguation r(t)= (3t+ 9) i + (8t - 2) j+(4t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Thenwrite the particle's velocity at t= 0 as a product of its speed and direction.What is the velocity vector?v(t) = (D i+ (Di+ k

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Description: The eguation r(t)= (3t+ 9) i + (8t - 2) j+(4t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Thenwrite the particle's velocity at t= 0 as a product of its speed and direction.What is the v

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Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter3: Motion In Two Dimensions

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The eguation r(t)= (3t+ 9) i + (8t - 2) j+(4t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then
write the particle's velocity at t= 0 as a product of its speed and direction.
What is the velocity vector?
v(t) = (D i+ (Di+ k

Expert Solution

Step 1

Introduction:

Kinematics helps to find out the velocity, acceleration of the particle at any time t when position of the particle is given. The relation between displacement, velocity and acceleration helps to find the unknown quantity.

Given data:

The position of the particle in the space at time t is,

$r (t) = (3 t + 9) \hat{i} + (8 t 2 - 2) \hat{j} + (4 t) \hat{k}$

Where,

t- time

i, j, k unit vectors in the direction of x, y, z axis

Formula used:

The relation between displacement is the, 'velocity is the time derivative of displacement'

$v = \frac{d r (t)}{d t}$

Similarly relation between displacement and acceleration is,

$a = \frac{d^{2} r (t)}{d t^{2}} = \frac{d v (t)}{d t}$

Step 2

Calculation:

Velocity of the particle is,

$v = \frac{d r (t)}{d t} v = \frac{d}{d t} ((3 t + 9) \hat{i} + (8 t^{2} - 2) \hat{j} + (4 t) \hat{k}) v = (3) \hat{i} + (16 t) \hat{j} + (4) \hat{k} Velocity at time t = 0 \sec is, v = (3) \hat{i} + (0) \hat{j} + (4) \hat{k} v = (3) \hat{i} + (4) \hat{k}$

Acceleration of the particle is,

$a = \frac{d^{2} r (t)}{d t^{2}} = \frac{d v (t)}{d t} a = \frac{d}{d t} ((3) \hat{i} + (16 t) \hat{j} + (4) \hat{k}) a = (0) \hat{i} + (16) \hat{j} + (0) \hat{k} a = (16) \hat{j} At time t = 0 \sec, acceleration of the particle is, a = (16) \hat{j}$

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The eguation r(t)= (3t+ 9) i + (8t - 2) j+(4t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. Then write the particle's velocity at t= 0 as a product of its speed and direction. What is the velocity vector? v(t) = (D i+ (Di+ k