BPSC (Bihar PCS) 2025 · Question 69 of 145
General Science
The truck covers 400 meters in 20 seconds from rest with constant acceleration. Given value, Distance travelled ( s) = 400 metre, Time = 20 sec Using the formula: F= m×a F= 7000 × a (1 tons= 1000 kg) Using the equation of motion (S= ut + ½ at2) 400 = ½ a × 400 a= 2 m/s2 F= 7000×2 = 14000N So, the force acting on the truck is 14,000 Newtons Therefore, option (D) is the correct answer.
Source: 71st BPSC CCE Prelims 2025 (Set H)
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