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

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 39
Followed by
Problem 41
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