Question

a police man is running towards a Thief with a constant acceleration of 0.1m/s2 . at a particular instant t=0, the policeman is 20 m behind the thief and at that velocity of policeman is 4 m/s. At this instant the thief observes the policeman coming towards him. So he starts running away from the policemen in the same direction with a constant acceleration. Then (i)What is the minimum constant acceleration with which the thief should run so that he is not caught? (a)0.2 m/s2 (b)0.4 m/s2 (c)0.5 m/s2 (d)0.8 m/s2 ​ ​(ii)if the thief runs with an acceleration 20% more than the minimum calculated above, then how close to the thiief can the policeman get? (a)2m (b)4m (c)6m(d)8m (iii)if the thief runs with an acceleration as considered in the previous question, then how far from the thief is the policeman when the rate at which the policeman approaches the thief is maximum? (a)20m (b)8m (c)12m (d)16m

Answer 1

Answer:

(i) c - 0.5m/s²

Explanation:

(i) For police

u = 4m/s  

a = 0.1 m/s^2  

For thief  

u = 0 ,  

Let time t velocity of police becomes Vp and of thief Vth for not to catch Vth is greater than or equal to Vp or their relative velocity becomes zero.  

u₍p₎⁽th⁾ = 4-0 = 4m/s  

ap(th) = ap - ath = (0.1 - a) m/s^2

s = 20 m  

Vp(th) = 0  

V^2 = u^2 +2as  

40 a = 20  

a = 0.5 m/s^2

I don't think I would be able to help with the rest of the questions but here is what I could :)

Answer 2

Answer:

option c 0.5 m/s^2 100% sure