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### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

# A racing car starts from rest at ${\mathbit{t}}{\mathbf{=}}{\mathbf{0}}$ and reaches a final speed ${\mathbit{v}}$ at time t . If the acceleration of the car is constant during this time, which of the following statements are true? (a) The car travels a distance ${\mathbit{v}}{\mathbit{t}}$ . (b) The average speed of the car is ${\mathbit{v}}{\mathbf{/}}{\mathbf{2}}$. (c) The magnitude of the acceleration of the car is ${\mathbit{v}}{\mathbf{/}}{\mathbit{t}}$ .(d) The velocity of the car remains constant. (e) None of statements (a) through (d) is true.

Hence, options (b) and (c) are correct.

See the step by step solution

## Step 1: Given value in the question

The acceleration of the car in the initial state is, ${v}_{0}=0 \text{m/s}$,

Final speed of the car $=v$

The expression used to calculate the acceleration of the car is given below.

role="math" localid="1663735439132" ${\mathbit{a}}{\mathbf{=}}\frac{\mathbf{v}\mathbf{-}{\mathbf{v}}_{\mathbf{0}}}{\mathbf{\Delta }\mathbf{t}}$

## Step 2: Substitution of the values in the expression

Put the given value in the expression is:

$a=\frac{v-0\text{m}/\text{s}}{t}\phantom{\rule{0ex}{0ex}}=\frac{v}{t}\phantom{\rule{0ex}{0ex}}$

So, option (C) is correct.

The displacement of the car to use the following kinetic equation will be

${v}^{2}={v}_{0}^{2}+2ax$

The above equation will be rearranged for x ,that is

$x=\frac{{v}^{2}-{v}_{0}^{2}}{2a}\phantom{\rule{0ex}{0ex}}x=\frac{{v}^{2}-{\left(0 \text{m/s}\right)}^{2}}{2\left(\frac{v}{t}\right)}\phantom{\rule{0ex}{0ex}}=\frac{vt}{2}\phantom{\rule{0ex}{0ex}}$

## Step 3: Determining the value of average speed

The average speed of the car is:

${v}_{avg}=\frac{x}{t}\phantom{\rule{0ex}{0ex}}=\frac{vt}{2t}\phantom{\rule{0ex}{0ex}}=\frac{v}{2}\phantom{\rule{0ex}{0ex}}$

So, option (b) is correct.