# Chapter 2 - Motion Along a Straight Line - Problems - Page 39: 108

$a=2.78m/s^2$

#### Work Step by Step

All units must be in SI units. For speed, the SI unit is $m/s$. Therefore, dimensional analysis must be used on the speed to get $$=\frac{360km}{1hr} \times \frac{1hr}{60min} \times \frac{1min}{60s} \times \frac{1000m}{1km}$$ $$=100m/s$$ To find the acceleration needed, use a kinematics equation relating distance, acceleration, initial velocity, and final velocity. This is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Substituting known values of $v_o=0.00m/s$, $v_f=100m/s$, and $\Delta x=1.80km(1000m/km)=1800m$ yields an acceleration of $$a=\frac{(100m/s)^2}{2(1800m)}$$ $$a=2.78m/s^2$$

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