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A thif is moving with a constant velocity of 10 m/s .200m ahead of policemen who chases him with a initial velocity of policemen 1 m/s and acceleration of 1 m/s² toward the thif how long will the policeman take to catch him

Answer 1

Answer:

30.93

Explanation:

Distance covered by the thief in time t

$S= v*t$

S= 10t       -   ----------1

To catch thief in time t,

Distance covered by policeman=

S+200m        ----------2

Also,

$S=ut+\frac{1}{2} at^{2}$

Therefore from eqn 2

S+200= $1t+\frac{1}{2}1t^{2}$

Substituting value of S from eqn 1

$10t+200= t+\frac{1}{2}t^{2} \\10t +200= \frac{2t+t2}{2}20t+400= 2t+t2\\ 18t+400=t2\\0=t2-18t-400\\Solving the eqn\\t= 30.93$

Therefore it will take him approx. 30.93 seconds

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