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

Answer 1

Firstly, we will find the distance between the theif and policeman.

⇒Usual speed = Speed of policeman - Speed of theif

⇒ Usual speed = 15 - 10

⇒Usual speed = 5 km/h

Now, we will change km/h into m/s

⇒Usual speed = 5 * (5/18)

⇒Usual speed = 25/18 m/s

___________________________

Distance covered = 400 m

Speed = 25/18 m/s

\Large{\star{\boxed{\sf{Time = \dfrac{Distance}{Speed} }}}}⋆

Time=

Speed

Distance

\begin{lgathered}\sf{\implies Time = \dfrac{400}{\dfrac{25}{18}}} \\ \\ \sf{\implies Time = \cancel{400} \times \frac{18}{\cancel{25}}} \\ \\ \sf{\implies Time = 16 \times 18} \\ \\ \sf{\implies Time = 288 \: seconds}\end{lgathered}

⟹Time=

18

25

400

⟹Time=

400

×

25

18

⟹Time=16×18

⟹Time=288seconds

Now, we have now time in which The policeman catch the thief.

Speed of theif = 10 km/h = 50/18 m/s

Time taken = 288 s

\Large{\star{\boxed{\sf{Distance = Speed \times Time}}}}⋆

Distance=Speed×Time

\begin{lgathered}\sf{Distance = \dfrac{50}{\cancel{18}} \times \cancel {288}} \\ \\ \sf{\implies Distance = 50 \times 16} \\ \\ \sf{\implies Distance = 800 \: m}\end{lgathered}

Distance=

18

50

×

288

⟹Distance=50×16

⟹Distance=800m

\Large{\star{\boxed{\sf{Distance = 800 \: m}}}}⋆

Distance=800m

∴ Option B is correct

Answer 2

a = 0.5 m/s^2

Explanation:

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 reltive velocity becomes zero .

upth = 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