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Grade: 9

                        

a police man is running toward a thief with a constant acceleration of 0.1 m/s^2. At a particular instant t=0,the policeman is 20 m behind the thief and at the instant the velocity of policeman is 4 m/s. At this instant the thief observed the policeman coming towards him. So he starts running away from the policeman in the same direction with a constant acceleration.Then what is the minimum acceleration with which the thief should run so that he is'nt caught

2 months ago

Answers : (1)

Anwar Sadath
28 Points
							
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
2 months ago
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