# 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...

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.

The following kinematic equations apply to situations involving * constant acceleration*,

*a*:

`v(t)=v_0+at` (A)

`x(t) = x_0+v_0t+(a/2)t^2` (B)

Let the acceleration be a constant value of `-3g=-3(9.8 m* s^-1)` . The car is stopped when *v* = 0 m/s, so use (A) to find the stopping time:

`0=v_0+at=(120 {km}/h)({1000 m}/{ 1 km})({1 h}/ {3600 s})-3(9.8 m*s^-2)*t`

`0=33.333 m/s-29.4t`

Solve for *t*: *t* = 1.135 s.

So, it takes 1.135 s to decelerate the car from 120 km/h to 0 km/h at a "3g" deceleration.

The stopping distance, `x - x_0` , can be found from (B) and by substituting the stopping time, *t* = 1.135 s:

`x-x_0=v_ot+(a/2)t^2=(120 {km}/h)({1000 m}/{ 1km})({1 h}/ {3600 s})(1.345 s)-(3/2)(9.8 m*s^-2)*(1.345 s)^2=18.89 m` The car travels 18.89 meters during the stopping time. The driver must start braking 18.89 m from the target stopping point.