# Chapter 2 - Motion Along a Straight Line - Problems: 23

$a=1.62\times 10^{13}m/s^2$

#### Work Step by Step

To find the acceleration using displacement, initial velocity, and final velocity, use the equation $$v_f^2=v_o^2-2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Substituting known values of $v_o=1.50\times 10^5m/s$, $v_f=5.70\times 10^6m/s$, and $\Delta x=1.00m$ yields an acceleration of $$a=\frac{(5.70\times 10^6m/s)^2-(1.50\times 10^5m/s)^2}{2(1.00m)}$$ $$a=1.62\times 10^{13}m/s^2$$

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