Question:
A car starts from rest and moves along the x-axis with constant acceleration $5 \mathrm{~m} \mathrm{~s}^{-2}$ for $8 \mathrm{~seconds}$. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest ?
Solution:
Here, $u=0, a=5 \mathrm{~m} \mathrm{~s}^{-2}, t=8 \mathrm{~s} .$ Distance travelled in first $8 \mathrm{~s}, x_{1}=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 5 \times 64=160 \mathrm{~m} .$
Velocity after $8 \mathrm{~s}, \mathrm{~V}=\mathrm{u}+\mathrm{at}=0+5 \times 8=40 \mathrm{~ms}^{-1}$.
Distance travelled in last $4 \mathrm{~s}$ moving with constant velocity, $x_{2}=v t=40 \times 4=160 \mathrm{~m}$
$\therefore$ Total distance $=x_{1}+x_{2}=320 \mathrm{~m}$.