At time $t = 0$ a drag racer starts from rest at the origin and moves along a straight line with velocity given by $v = 5t^2$, where $v$ is in $\text{m/s}$ and $t$ in $\text{s}$.  The expression for the displacement of the car from $t = 0$ to time $t$ is

$ \textbf{(A)}\ 5t^3 \qquad\textbf{(B)}\ 5t^3/3\qquad\textbf{(C)}\ 10t \qquad\textbf{(D)}\ 15t^2 \qquad\textbf{(E)}\ 5t/2 $
1. The velocity of the drag racer is given by \( v = 5t^2 \) m/s.
2. To find the displacement \( x \) from \( t = 0 \) to time \( t \), we need to integrate the velocity function with respect to time. The displacement \( x \) is given by:
   \[
   x = \int_0^t v(t') \, dt'
   \]
3. Substituting the given velocity function \( v(t') = 5t'^2 \) into the integral, we get:
   \[
   x = \int_0^t 5t'^2 \, dt'
   \]
4. To solve the integral, we use the power rule for integration, which states that \( \int t'^n \, dt' = \frac{t'^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Here, \( n = 2 \):
   \[
   x = 5 \int_0^t t'^2 \, dt' = 5 \left[ \frac{t'^3}{3} \right]_0^t
   \]
5. Evaluating the definite integral from \( 0 \) to \( t \):
   \[
   x = 5 \left( \frac{t^3}{3} - \frac{0^3}{3} \right) = 5 \cdot \frac{t^3}{3} = \frac{5t^3}{3}
   \]

Conclusion:
\[
\boxed{\frac{5t^3}{3}}
\]