Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 63: 48

The car will stop before hitting the child.

$40km/hr=40\times\frac{1000m}{3600s}=11.11m/s$ distance traversed in first 0.25 s = $11.11\times 0.25 m=2.78m$ Remaining distance $=13-2.78=10.22m$ $v^{2}=u^{2}+2aS$ $0=11.11^{2}-2\times8\times S$ $S=7.71m$ This is less than 10.22m, hence the car will stop before hitting the child.