## Fundamentals of Physics Extended (10th Edition)

Published by Wiley

# Chapter 2 - Motion Along a Straight Line - Problems: 114a

#### Answer

$9.1m/s^2$

#### Work Step by Step

All units must be in SI units. The SI unit for speed is m/s. Therefore, the speed must be converted to m/s using dimensional analysis. $$\frac{200km}{1hr} \times \frac{1hr}{60min} \times \frac{1min}{60s} \times \frac{1000m}{1km}$$ $$=55.6m/s$$ To find the acceleration, use the kinematics equation relating acceleration, distance, initial velocity, and final velocity. This equation is $$v_f^2=v_o^2+2a\Delta x$$ Since the object ends at rest, $v_f=0.00m/s$. $$0=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=-\frac{v_o^2}{2\Delta x}$$ Substituting known values of $v=55.6m/s$ and $\Delta x=170m$ yields an acceleration of $$a=-\frac{(55.6m/s)^2}{2(170m)}=-9.1m/s^2$$ The magnitude of the acceleration is always positive, so the magnitude must be $9.1m/s^2$.

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