Final answer:
To catch up with a cockroach that scurries away at 1.50 m/s while you approach it at 0.70 m/s, the minimum constant acceleration you need to achieve over a total distance of 2.05 m (accounting for your initial speed) in the time it takes for the cockroach to travel 1.30 m is 2.00 m/s².
Step-by-step explanation:
The question involves a scenario where you need to calculate the minimum constant acceleration required to catch up with a cockroach. First, we need to find out how long it takes for the cockroach to cover the 1.30 m distance. Since the cockroach's speed is constant (1.50 m/s), we can find the time (t) using the formula:
distance = speed × time
1.30 m = 1.50 m/s × t
So, t = 1.30 m / 1.50 m/s = 0.867 s.
You start 0.75 m behind and approach the cockroach at 0.70 m/s, so in that time, you will cover:
distance = speed × time
= 0.70 m/s × 0.867 s
= 0.608 m
You need to cover not just the 0.75 m you start behind, but also an additional 1.30 m the cockroach travels. So you need to cover 2.05 m total (1.30 m + 0.75m - 0.608m), but since you are already moving at 0.70 m/s, the actual distance you need to 'accelerate' across is:
2.05 m - 0.608 m = 1.442 m
Let's denote the required acceleration as 'a'. To find this acceleration, we can use the kinematic equation:
d = v₀t + 0.5at²
Where 'd' is the distance to accelerate across (1.442 m), 'v₀' is your initial velocity (0.70 m/s), and 't' is the time calculated earlier (0.867 s). Rearranging the equation for 'a', we get:
a = (2(d - v₀t)) / t²
= 2(1.442 m - (0.70 m/s × 0.867 s)) / (0.867 s)²
= 2.00 m/s²
Therefore, the minimum constant acceleration needed is 2.00 m/s².