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Expert-verified Found in: Page 32 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # The following equations give the velocity of a particle in four situations: ${\mathbf{\left(}}{\mathbf{a}}{\mathbf{\right)}}{\mathbit{v}}{\mathbf{=}}{\mathbf{3}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{b}}{\mathbf{\right)}}{\mathbit{v}}{\mathbf{=}}{\mathbf{4}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{t}}{\mathbf{-}}{\mathbf{6}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{c}}{\mathbf{\right)}}{\mathbit{v}}{\mathbf{=}}{\mathbf{3}}{\mathbit{t}}{\mathbf{-}}{\mathbf{4}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{d}}{\mathbf{\right)}}{\mathbit{v}}{\mathbf{=}}{\mathbf{5}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{3}}$ .To which of these situations do the equations of Table 2-1 apply? For situations a) and c) the equations of Table 2.1 can be applied.

See the step by step solution

## Step 1: Given information

Table 2.1 with equations of motion with constant acceleration.

## Step 2: Acceleration

The problem uses the simple concept of acceleration which is the rate of change of velocity with time. The general formula for acceleration can be used to find the situations to which the equations in table 2.1 can be applied.

Formula:

Acceleration (a)$=\frac{dv}{dt}$

## Step 3: To find the situations in which the equations in table 2.1 can be applied

The equations of table 2.1 only apply to the situations in which acceleration is constant.

Acceleration can be defined as,

Acceleration (a)$=\frac{dv}{dt}$

(a) The given velocity fuction is ,v = 3.so,

$a=\frac{d\left(3\right)}{dt}\phantom{\rule{0ex}{0ex}}=0$

Acceleration is constant in this situation.

Therefore, the equations of Table 2.1 can be applied to (a).

(b) The given velocity function is,

role="math" localid="1656997781414" $v=4{t}^{2}+2t-6$

So

role="math" localid="1656997790862" $v=\frac{d\left(4{t}^{2}+2t-6\right)}{dt}\phantom{\rule{0ex}{0ex}}=8t+2$

Acceleration is not constant in this situation.

Therefore, the equations of table 2.1 cannot be applied to (b).

(c)So,

$a=\frac{d\left(3t-4\right)}{dt}\phantom{\rule{0ex}{0ex}}=3$

Acceleration is constant in this situation.

Therefore, the equations of Table 2.1 can be applied to (c).

(d) The given velocity function is,

$v=5{t}^{2}-3$ So,

$a=\frac{d\left(5{t}^{2}-3\right)}{dt}\phantom{\rule{0ex}{0ex}}=10t$

Acceleration is not constant in this situation.

Therefore, the equations of table 2.1 cannot be applied to (d). ### Want to see more solutions like these? 