Question

Large cockroaches can run as fast as 1.50 m/s in short bursts. suppose that you turn on the light in a cheap motel and see a cockroach scurrying away from you in a straight line at a constant 1.50 m/s as you move toward it at 0.70 m/s. if you start 0.75 m behind it, what minimum constant acceleration would you need to catch up with the cockroach when it has traveled 1.30 m, just short of safety under a counter?

Answer 1

In order to catch up with the cockroach, the time it took for it to reach just before under the counter must be the same time for me to travel its covered distance. Thus,

time of cockroach = time of my motion
Time of cockroach = Distance/Constant Speed = 1.30 m/1.50 m/s = 0.867 seconds

We use this for the equation below:
x = v₀t + 0.5at²
0.75 m = (0.7 m/s)(0.867 s) + 0.5(a)(0.867 s)²
Solving for a,
a = 0.381 m/s²