Answer to Question #133567 in Mechanics | Relativity for Brooke

Question

A car traveling at a constant speed of 32.5m/s squares passes a trooper hidden behind a billboard.1 second later the trooper starts the car with a constant acceleration Of 3.8m/s squared.how long after the trooper starts the chase.answer in units of s

Answer 1

Let us write the equations of motion of the objects. x = 0 at the billboard, t=0 when the car passes a billboard.

For a car $x_1(t) = v\cdot t,$ for the trooper $x_2(t) = 0 + 0\cdot(t-1) + \dfrac{a(t-1)^2}{2}$ (only when t > 1). Here t is in seconds.

When the trooper reaches the car $x_1 = x_2, \;\;\; vt = \dfrac{a(t-1)^2}{2}$ ,

$at^2-t(2a+2v) + a = 0.$

$3.8t^2 - t(2\cdot3.8 + 2\cdot32.5) + 3.8 = 0, \;\; 3.8t^2 - 72.6t + 3.8 = 0.$

We may solve this equation as a standard quadratic equation and get $t_1\approx 0.05\,\mathrm{s}, \; t_2\approx 19.05\,\mathrm{s}.$

The first root is less than 1 second, so we should choose the second root.