Question

A burglar's car had started with an acceleration of 2 m/s A police vigilant party came after 5s
at the same position and continued to chase the burglar's car with a uniform velocity of 20 m/s.
Find the time taken in which the police van will overtake the burglar's car​

Answer 1

Answer

5 sec

Explanation:

Burgular's Car (Given)

Initial velocity = 0 m/s

acceleration = 2 m/s²

Police Vigilant Party (Given)

Velocity = 20 m/s

Now,

If the police comes after 5 seconds,

then,

Distance covered by burgular's Car in 5 seconds will be,

\begin{gathered}s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times t + \frac{1}{2} \times 2 \times {5}^{2} \\ s = 0 + 25\\ s = 25 \: metres\end{gathered}

s=ut+

2

1

at

2

s=0×t+

2

1

×2×5

2

s=0+25

s=25metres

Now,

Let the time taken to over take the burgular's Car be x

then,

Distance covered by police = Distance covered by the Burgular's Car + 25.

(v × t) = (ut + \frac{1}{2} a {t}^{2} ) + 25(ut+

2

1

at

2

)+25

(20 * x) = (10 \times x + \frac{1}{2} \times 2 \times {x}^{2} ) + 25(10×x+

2

1

×2×x

2

)+25

20x = x² +10x + 25

0 = x² + 10x - 20x + 25

0 = x² - 10x + 25

0 = x² - 5x - 5x +25

0 = x(x-5) -5(x-5)

0 = (x-5) (x-5)

x = 5 seconds.

After 5 seconds the police car will overtake the Burgular's Car.

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