Question

need help on the attachment @Calculus1

Answer 1

\[d=\int\limits vdt\]
But in this case it's easier to think of this integral as the area under the graph.

Looks like at t=3 we have a triangle from t=0 to t=1
the area of this triangle is 1

from t=1 to t=3 we seem to have a square
the area of this square is 4

The total area from t=0 to t=3 is then 5

Since the v-coordinates are in multiples of 10 yards/sec, the total distance is
10(5)=50yards