Fixed formatting and TeX errors in solution

Author: paulvojta

OpenProblemLibrary/ma122DB/set10/s4_10_75.pg

@@ -44,7 +44,7 @@ Answer: \{ans_rule(30) \}
 END_TEXT
 
 $t1 = -$v1/32;
-$ans =   "$v1*$v1/64";
+$ans =   $v1*$v1/64;
 ANS(num_cmp($ans));
 
 
@@ -55,10 +55,10 @@ position at time \( t \) (\( s(t) \)), instantaneous velocity
 at time \( t \) (\( v(t) \)), and acceleration at time \( t\) (\(a(t)\)).
 This relationship is given by:
 $BR$BR
-\[  \begin{array}
-    v(t)  = s'(t) \\
-    a(t)  = v'(t) = s''(t) \\
-  \end{array}
+\[  \begin{split}
+    v(t) &= s'(t) \\
+    a(t) &= v'(t) = s''(t) \\
+  \end{split}
 \]
 $BR$BR
 With this in mind, this problem becomes an exercise in finding
@@ -92,11 +92,11 @@ and we know the stone struck the ground (so its position was 0) at
 \( t = !{$t1:%5.3f} \), we can solve the following equation for \( C_2 \).
 $BR$BR
 \[
-  \begin{array}
-    s(!{$t1:%5.3f}) =  0 \\
-    -16(!{$t1:%5.3f})^2 + C_2 = 0 \\
-    C_2 = !{$ans:%5.3f} \\
-  \end{array}
+  \begin{split}
+    s(!{$t1:%5.3f}) &= 0 \\
+    -16(!{$t1:%5.3f})^2 + C_2 &= 0 \\
+    C_2 &= !{$ans:%5.3f} \\
+  \end{split}
 \]
 Now, since we want to know the height from which the stone was dropped, we 
 just need to find its position at time \( t = 0 \).  But this is \( s(0) = -16(0)^2+ $ans \)