Answer
(a) $\Delta x=10~m$
(b) $\Delta x=20~m$
(c) $\Delta x=40~m$
Work Step by Step
Constant-acceleration motion:
$v_{av}=\frac{v_{0}+v}{2}$. See equation 2-9, page 32.
And $v_{av}=\frac{\Delta x}{\Delta t}$. Rearranging the equation:
$\Delta x=(v_{av})(\Delta t)$
(a) The segment A begins with $v_{0}=0,~t_{0}=0$ and finishes with $v_{1}=2~m/s,~t_{1}=10~s$. So, $\Delta t=t_{1}-t_{0}=10~s-0=10~s$.
$v_{av}=\frac{v_{0}+v_{1}}{2}=\frac{0+2~m/s}{2}=1~m/s$
$\Delta x=(v_{av})(\Delta t)=(1~m/s)(10~s)=10~m$
(b) The segment B begins with $v_{1}=2~m/s,~t_{1}=10~s$ and finishes with $v_{2}=6~m/s,~t_{2}=15~s$. So, $\Delta t=t_{2}-t_{1}=15~s-10~s=5~s$.
$v_{av}=\frac{v_{1}+v_{2}}{2}=\frac{2~m/s+6~m/s}{2}=4~m/s$
$\Delta x=(v_{av})(\Delta t)=(4~m/s)(5~s)=20~m$
(c) The segment C begins with $v_{2}=6~m/s,~t_{2}=15~s$ and finishes with $v_{3}=2~m/s,~t_{3}=25~s$. So, $\Delta t=t_{3}-t_{2}=25~s-15~s=10~s$.
$v_{av}=\frac{v_{2}+v_{3}}{2}=\frac{6~m/s+2~m/s}{2}=4~m/s$
$\Delta x=(v_{av})(\Delta t)=(4~m/s)(10~s)=40~m$
