Answer
Running at a constant velocity would take 2.81 s longer
Work Step by Step
$x=ut+\frac{1}{2}at^{2}$
When a = 0.0105
$(1609)=(16.58)t+\frac{1}{2}(0.0105)t^{2}$
$0.00525t^{2}+16.58t-1609=0$
Using the quadratic formula:
$t=\frac{-(16.58)±\sqrt ((16.58)^{2}-4(0.00525)(-1609)}{2(0.00525)}$
$t\approx 94.23, t\approx -3252.33$
$t\approx 94.23\;s$ since $t\gt0$
When a = 0:
$(1609)=(16.58)t+\frac{1}{2}(0)t^{2}$
$t=\frac{1609}{16.58}\approx 97.04\;s$
97.04 - 94.23 = 2.81 s
Hence, running at a constant velocity would take 2.81 s longer
