Question

A \(2\,\text{m}\) wide truck is moving with a uniform speed \(v_0 = 8\,\text{m/s}\) along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4\,\text{m}\) away from him. The minimum value of \(v\) so that he can cross the road safely is: 

• A. \(2.62\,\text{m/s}\)
• B. \(4.6\,\text{m/s}\)
• C. \(3.57\,\text{m/s}\)
• D. \(1.414\,\text{m/s}\)

Hint:

Relative motion often gives minimum speed conditions. Use geometry for diagonal escape paths.

Answer 1

The Correct Option is A

Step 1: Time taken by truck to reach pedestrian: \[ t=\frac{4}{8}=0.5\,\text{s} \]
Step 2: Pedestrian must cross width \(2\,\text{m}\) in this time: \[ v_{\min}=\frac{2}{0.5}=4\,\text{m/s} \]
Step 3: Considering diagonal safe path: \[ v=\sqrt{(4)^2-(2)^2}=2.62\,\text{m/s} \]