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The front 1. 00 m of a 1400-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. If a car traveling 25. 0 m/s stops uniformly in 1. 00 m,

Respuesta :

The acceleration of the car in terms of g will be -26.6 g. Newton's second equation of the motion is used in the given problem.

What is acceleration?

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The given data in the problem is;

From the Newtons third equation of motion;

[tex]\rm v^2=u^2+2as\\\\ a=\frac{v^2-u^2}{2s} \\\\ a=\frac{0^2-25 ^2}{2 \times 1.20\ }[/tex]

[tex]\rm a= -260.41 m/s^2 \\\\ a= \frac{-260.41}{9.81} \\\\ a = - 26.573129 \ g \\\\ a=- 26.6 \ g[/tex]

Hence,the acceleration of the car in terms of g will be -26.6 g.

To learn more about acceleration, refer to the link;

https://brainly.com/question/2437624'

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