A car starts from rest and moves along the $x-$ axis with constant acceleration $5\, ms^{-2}$ for $8\,\sec $. If it then continues with constant velocity, what distance will the car cover in $12\,\sec $ since it started from the rest ?
A car starts from rest and moves along the $x-$ axis with constant acceleration $5\, ms^{-2}$ for $8\,\sec $. If it then continues with constant velocity, what distance will the car cover in $12\,\sec $ since it started from the rest ?
The distance travelled in first $8 \,s ,\,\, x_{1}=0+\frac{1}{2}(5)(8)^{2}=160\, m$
At this point the velocity $v=u+a t=0+(5 \times 8)=40 \,m s^{-1}$
Therefore, the distance covered in last four seconds, $x _{2}=(40 \times 4)\, m =160\, m$
Thus, the total distance $x=x_{1}+x_{2}=(160+160) \,m =320\, m$
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