Question

A thief 40 meters away runs towards you at 8m/s. What is the smallest acceleration so that v=at keeps you in front?

The answer is 8/5 m/s^2. However, I have tried 1 m/s^2 (which is smaller than the answer) and it still can keep the person in front...

Many thanks!

Answer 1

The thief's target must reach a speed of 8 m/s in t seconds. Assuming constant acceleration, the average speed of the target during acceleration is 8/2 = 4 m/s.
The distance traveled by the target during the acceleration period is 4t.
The distance traveled by the thief while his target is accelerating is (40 + 4t) m.
The time taken by the thief to just reach his target is:
\[\frac{40+4t}{8}=5+0.5t\ seconds\]
But this time is equal to the time for the target to accelerate to 8 m/s. Therefore:
\[5+0.5t=t\]
Solving gives the value for t as 10 seconds.
Substituting in v = at we get:
8 = 10a
Therefore the minimum acceleration is 0.8 m/s^2