The velocity of a projectile ﬁred straight into the air is given every half second. Use right endpoints to estimate the distance the projectile travelled in four seconds. $$\begin{array} { c | c }
\begin{array} { c }
\text { Time } \
( \mathrm { sec } )
\end{array} & \begin{array} { c }
\text { Velocity } \
( \mathrm { m } / \mathrm { sec } )
\end{array} \
\hline 0 & 140 \
0.5 & 135.1 \
1.0 & 130.2 \
1.5 & 125.3 \
2.0 & 120.4 \
2.5 & 115.5 \
3.0 & 110.6 \
3.5 & 105.7 \
4.0 & 100.8
\end{array}$$
A) 471.8 m
B) 982.8 m
C) 491.4 m
D) 943.6 m

Question

The velocity of a projectile ﬁred straight into the air is given every half second. Use right endpoints to estimate the distance the projectile travelled in four seconds. $$\begin{array} { c | c } 
\begin{array} { c } 
\text { Time } \
( \mathrm { sec } )
\end{array} & \begin{array} { c } 
\text { Velocity } \
( \mathrm { m } / \mathrm { sec } )
\end{array} \
\hline 0 & 140 \
0.5 & 135.1 \
1.0 & 130.2 \
1.5 & 125.3 \
2.0 & 120.4 \
2.5 & 115.5 \
3.0 & 110.6 \
3.5 & 105.7 \
4.0 & 100.8
\end{array}$$
A) 471.8 m
B) 982.8 m
C) 491.4 m
D) 943.6 m

Quizplus · Accepted Answer

The distance traveled is the area under the velocity-time graph. Using right endpoints, the area is approximated by the sum of the areas of the rectangles with heights equal to the velocities at the right endpoints and widths of 0.5 seconds. The area of the first rectangle is 140 * 0.5 = 70 m, the area of the second rectangle is 135.1 * 0.5 = 67.55 m, and so on. The total area is 70 + 67.55 + 65.1 + 62.65 + 60.2 + 57.75 + 55.3 + 52.85 + 50.4 = 471.8 m.

The Velocity of a Projectile ﬁRed Straight into the Air