A1000kg vehicle moving with a speed of 10 m/s is brought to rest at distance of 50m by application if breaks find acceleration and the force acting the vehicle
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a = -u² / (2S) ………[∵ v = 0]
= -(10 m/s)² / (2 × 50 m)
= -100 m/s²
Acceleration is -100 m/s². Here negative sign indicates that acceleration of vehicle is in opposite direction of its motion.
F = ma
= (1000 kg) × (-100 m/s²)
= -10⁵ N
Force acting on vehicle is -10⁵ N. Here negative sign indicates that Force acting on vehicle is in opposite direction of its motion.
= -(10 m/s)² / (2 × 50 m)
= -100 m/s²
Acceleration is -100 m/s². Here negative sign indicates that acceleration of vehicle is in opposite direction of its motion.
F = ma
= (1000 kg) × (-100 m/s²)
= -10⁵ N
Force acting on vehicle is -10⁵ N. Here negative sign indicates that Force acting on vehicle is in opposite direction of its motion.
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