Suggested languages for you:

Americas

Europe

Q2OQ

Expert-verifiedFound in: Page 49

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**A racing car starts from rest at ${\mathit{t}}{\mathbf{=}}{\mathbf{0}}$** **and reaches a final speed ** ${\mathit{v}}$** 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 ${\mathit{v}}{\mathit{t}}$** **. **

**(b) The average speed of the car is ${\mathit{v}}{\mathbf{/}}{\mathbf{2}}$****. **

**(c) The magnitude of the acceleration of the car is ${\mathit{v}}{\mathbf{/}}{\mathit{t}}$** **.**

**(d) The velocity of the car remains constant. **

**(e) None of statements (a) through (d) is true.**

** **

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

The acceleration of the car in the initial state is, ${v}_{0}=0\u200a\text{m/s}$,

Final speed of the car $=v$

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

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

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

$x=\frac{{v}^{2}-{v}_{0}^{2}}{2a}\phantom{\rule{0ex}{0ex}}x=\frac{{v}^{2}-{(0\u200a\text{m/s})}^{2}}{2\left(\frac{v}{t}\right)}\phantom{\rule{0ex}{0ex}}=\frac{vt}{2}\phantom{\rule{0ex}{0ex}}$

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.

94% of StudySmarter users get better grades.

Sign up for free