Question

$A2$ m wide truck is moving with a uniform speed of $8m/s$ along a straight horizontal road. A pedestrian starts crossing the road at an instant when the truck is $4m$ away from him. The minimum constant velocity with which he should run to avoid an accident is :

• A. $1.6\sqrt{5}m/s$
• B. $1.2\sqrt{5}m/s$
• C. $1.2\sqrt{7}m/s$
• D. $1.6\sqrt{7}m/s$

Answer 1

Answer: (A)

$(v\sin \theta )t=2$
$(8-v\cos \theta )t=4$
$\frac{v\sin \theta }{8-v\cos \theta }=\frac{1}{2}$
$2v\sin \theta =8-v\cos \theta$
$2v\sin \theta +v\cos \theta =8$
$v=\frac{8}{2\sin \theta +\cos \theta }$
$v_{\min }=\frac{8}{\sqrt{2^2+1^2}}$
$v_{\min }=\frac{8}{\sqrt{5}}$
$=\frac{8\sqrt{5}}{\sqrt{5}\sqrt{5}}$
$=1.6\sqrt{5}$