## Chemistry (7th Edition)

We find: Kinetic energy $K.E=\frac{1}{2}mv^{2}$ For the car, $K.E=\frac{1}{2}\times1400\times(115)^{2}=9.3\times10^{6}$ For the truck, $K.E=\frac{1}{2}\times12000\times(38)^{2}=8.7\times10^{6}$ Kinetic energy is more for the car.