Question

Attempt question no 12 after reading passage

PASSAGE -IV

A thief is running on a motorcycle at a constant speed of 25 ms–1. A police jeep starts chasing from a point 1.25 km behind him a uniform acceleration of 2 ms^–2.

Q12. What should be the minimum acceleration of the thief to escape from police.

(A) 1 ms^–2 (b) 1.25 ms^–2

(C) 1.5 ms^–2 (d) 1.75 ms^–2

Answer 1

Dear Student ,
Here in this case , the thief is moving with constant speed 25 m/s .
Now the policeman starts chasing from 1.25 km behind him and with acceleration of 2 m/s² .
$$\begin{gathered} Sofromtheequationofmotiontimetakenbythepolicemantoreachthethiefis, \\ S+ut=\frac{1}{2}a{t}^2 \\ or,1250+25t=\frac{1}{2}\times 2\times t^2 \\ or,t^2-25t-1250=0 \\ or,t^2-50t+25t-1250=0 \\ or,\left(t-50\right)\left(t+25\right)=0 \\ or,t=50s \\ Sovelocityofthepolice=v=u+at=0+\left(2\times 50\right)=100m/s \\ So,theminimumacceleartionofthethieftoescapefrompoliceis, \\ {a}_{min}=\frac{100-25}{50}=\frac{75}{50}=\frac{3}{2}=1.5m/s^2 \end{gathered}$$
Regards