1956 AHSME Problems/Problem 40

Problem 40

If $V = gt + V_0$ and $S = \frac {1}{2}gt^2 + V_0t$ , then $t$ equals:

$\textbf{(A)}\ \frac{2S}{V+V_0}\qquad \textbf{(B)}\ \frac{2S}{V-V_0}\qquad \textbf{(C)}\ \frac{2S}{V_0-V}\qquad \textbf{(D)}\ \frac{2S}{V}\qquad \textbf{(E)}\ 2S-V$

Solution

Solve for $t$ in the first equation: $V - V_0 = gt$ $t = \frac{V - V_0}{g}$ Then use the second equation to solve for $g:$ $2(S - V_0 t) = gt^2$ $g = \frac{2(S - V_0 t)}{t^2}$

Plug in $g$ in $t$ to get $\boxed{\textbf{(A)}}.$ -coolmath34

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 41
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1956 AHSME Problems/Problem 40

Problem 40

Solution

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