## Essential University Physics: Volume 1 (4th Edition)

We know that $v=\frac{9000+100}{35\times 60}$ $v=4.333\frac{m}{s}$ The leader arrives at the following time $t=\frac{900}{4.333}=208s$ Now we can determine acceleration as follows $x=x_{\circ}+v_{\circ}t+\frac{1}{2}at^2$ We plug in the known values to obtain: $1000=0+4.333(208)+\frac{1}{2}(a)(208)^2$ $a=0.0045\frac{m}{s^2}$