## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can convert the initial speed to units of m/s $v = (35~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$ $v = 9.7~m/s$ We can use conservation of energy to find the speed at the bottom of the hill. $KE_f+PE_f = KE_0+PE_0$ $\frac{1}{2}mv_f^2+0=\frac{1}{2}mv_0^2+mgh$ $v_f^2=v_0^2+2gh$ $v_f= \sqrt{v_0^2+2gh}$ $v_f= \sqrt{(9.7~m/s)^2+(2)(9.80~m/s^2)(15~m)}$ $v_f = 19.7~m/s$ We can convert the final speed to units of km/h $v = (19.7~m/s)(\frac{1~km}{1000~m})(\frac{3600~s}{1~hr})$ $v = 70.9~km/h$ The speed at the bottom of the hill is 70.9 km/h. Since this speed is over the speed limit of 70 km/h, it is possible that the police officer will give us a speeding ticket.