Answer to Question #86992 in Mechanics | Relativity for Ashlin

Question

A rabbit and a dog catch sight of each other when they are 2.7 m apart. The dog begins to chase the rabbit as the rabbit runs directly towards its burrow. The rabbit starts of 11 m from its burrow.

Both the dog and the rabbit move along the same straight line and both start from rest and accelerate at a constant rate. If the dog can accelerate at a rate of 3.3 m s−2 then at what rate is the rabbit required to accelerate in order to prevent being caught? (in m s−2 to 2 s.f)

Answer 1

As long as both the rabbit and the dog start from rest and move along the straight line, the condition for the rabbit to prevent being caught should be the following:

$$t_D \geq t_R$$

where t_D and t_R is the time needed to reach the burrow for the dog and the rabbit correspondingly (if the dog is capable of getting earlier to the burrow, then it surely intercepts the rabbit on its way).

The times are connected with the distances in this case via the following expressions:

$$L+\Delta l = \frac{a_D t_D^2}{2}, \\ L = \frac{a_R t_R^2}{2},$$

where L is the distance between the rabbit and the burrow, (delta l) is the initial separation between the dog and the rabbit, a_D and a_R are the accelerations of the dog and the rabbit correspondingly.

Hence, in the view of the inequality written above, we obtain:

$$\frac{L + \Delta l}{a_D} \geq \frac{L}{a_R}$$

Finally,

$$a_R \geq a_D \frac{L}{L+\Delta l}$$

Substituting the numerical values, we obtain:

$$a_R \geq 3.3 \frac{11}{11+2.7} \approx 2.6 \, m/s^2$$

Answer: the acceleration should be at least 2.6 m/s²