A driver cannot withstand any acceleration of over 3 g. If this is the case, what is the distance from a point where the driver needs to stop that he should apply his brakes if he is going at 120 km/h.

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The driver of a car cannot withstand an acceleration of more that 3g. g is the acceleration due to gravity and equal to 9.8 m/s^2; 3g is an acceleration equal to 29.4 m/^2.

If the driver is initially traveling at 120 km/h and needs to stop at a point. The deceleration when the brakes are applied cannot exceed 29.4 m/s^2. Let the distance where the brakes are applied from the point where he intends to stop be S m.

Using the formula `v^2 - u^2 = 2*a*s`

`0^2 - (120*(1000/3600))^2 = -2*29.4*S`

=> S = 18.89 m

The brakes should be applied at least 18.89 m from the point where the driver wishes to stop.

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