Question

A thief is running at the speed of 100 m/min. A policeman started running after him 1 min later with an acceleration of 10m/min. Find the time taken by the policeman to catch the thief.

Answer 1

Answer:

!!

Step-by-step explanation:

Let the police catch the thief in ‘n’ minutes

Since the thief ran 1 min before the police start running.

Time taken by the thief before he was caught = (n + 1) min

Distance travelled by the thief in (n+1) min = 100(n+1)m

Given speed of policeman increased by 10m per minute.

Speed of police in the 1st min = 100m/min

Speed of police in the 2nd min = 110m/min

Speed of police in the 3rd min = 120m/min

Hence 100, 110, 120… are in AP

Total distance travelled by the police in n minutes =(n/2)[2a+(n – 1)d]

= (n/2)[2 x 100 +(n – 1)10]

After the thief was caught by the police,

Distance traveled by the thief = distance travelled by the police

100(n+1)= (n/2)[2 x 100 +(n-1)10]

200(n + 1) = n[200 + 10n – 10]

200n + 200= 200n + n(10n – 10 )

200 = n(n – 1)10

n(n – 1) – 20 = 0

n2 – n– 20 = 0

n2 – 5n + 4n – 20 = 0

n(n – 5) + 4(n – 5) = 0

(n – 5) (n+4) = 0

(n – 5) = 0 or (n + 4) = 0

n= 5 or n= -4

Hence n= 5 since n cannot be negative

Therefore the time taken by the policeman to catch the thief = 5minutes

Answer 2

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5 minutes

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step-by-step explanation :

Let,

the police catch the thief in ‘t’ minutes

Since,

the thief ran 1 min before the police start running,

therefore,

Time taken by the thief before he was caught = (t + 1) min

Now,

we know that,

Distance = speed × time

therefore,

Distance travelled by the thief in (t+1) min

= 100(t+1)m

now,

Given,

speed of policeman increased by 10 m/min

therefore,

Speed of policeman in the 1st min = 100 m/min

again,

Speed of policeman in the 2nd min = 110 m/min

and

Speed of policeman in the 3rd min = 120 m/min

Hence,

on observing the pattern,

we get, an A.P i.e.,

100, 110, 120….............

where,

a = 100

d = 110 - 100 = 10

nOw,

By the formula for sum of n terms of an AP,

we get,

Total distance travelled by the police in t minutes

=(t/2)[2a+(t – 1)d]

Putting the value of a, and d,

we get,

= (t/2)[2 x 100 +(t – 1)10]

now,

After the thief was caught by the police,

Distance traveled by the thief = distance travelled by the police

=> 100(t+1)= (t/2)[2 x 100 +(t-1)10]

=> 200(t + 1) = t[200 + 10t – 10]

=> 200t + 200= 200t + t(10t – 10 )

=> 200 = t(t – 1)10

=> t(t – 1) – 20 = 0

=> ${t}^{2}$ – t– 20 = 0

=> ${t}^{2}$– 5t + 4t – 20 = 0

=> t(t – 5) + 4(t – 5) = 0

after factorisation,

we get,

=> (t – 5) (t +4) = 0

therefore,

(t – 5) = 0 or (t + 4) = 0

=> t = 5 or t = -4

but

we know that,

time can't be negative,

therefore,

t = 5

Hence,

the time taken by the policeman to catch the thief = 5 minutes