A vehicle of mass 900 kg moves along a straight horizontal road. At time \(t\) seconds the resultant force acting on the vehicle has magnitude \(\frac { 63000 } { k t ^ { 2 } } \mathrm {~N}\), where \(k\) is a positive constant. The force acts in the direction of motion of the vehicle. At time \(t\) seconds, \(t \geqslant 1\), the speed of the vehicle is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the vehicle is a distance \(x\) metres from a fixed point \(O\) on the road. When \(t = 1\) the vehicle is at rest at \(O\) and when \(t = 4\) the speed of the vehicle is \(10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Show that \(v = 14 - \frac { 14 } { t }\)
Hence deduce that the speed of the vehicle never reaches \(14 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Use the trapezium rule, with 4 equal intervals, to estimate the value of \(x\) when \(v = 7\)