# A car is traveling at 150 km/h. It comes to a crossroad and has to stop. The driver applies brakes at a distance of 50 m from the place where he has to stop and is just able to manage. What is the...

A car is traveling at 150 km/h. It comes to a crossroad and has to stop. The driver applies brakes at a distance of 50 m from the place where he has to stop and is just able to manage. What is the acceleration due to the application of the brakes.

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The initial speed of the car is 150 km/h. As the driver approaches a crossroad he realizes that he has to stop and applies his brakes at a point 50 m away from the point where he needs to stop. By doing so he is able to stop his car in time.

To determine the resulting acceleration due to the application of the brakes use the formula `v^2 - u^2 = 2*a*s` where v is the final velocity, u is the initial velocity and a is the acceleration of the object over the distance s.

Here, v = 0, u = 150, s = 50/1000 km

`0 - 150^2 = 2*a*(50/1000)`

=> `a = -150^2*10`

=> `a = -225000`

The application of the brakes decreases the speed of the car at a rate 225000 km/h^2. The acceleration here is negative and equal to -225000 km/h^2

To make life easy, assume the acceleration is constant. In this case, the kinematic equations of motion are applicable:

`v(t) =v_o + a*t` (A)

`x(t) = x_0+v_0*t+(a/2)t^2` (B)

The stopping distance, `x - x_0` , is 50 m and the initial speed is `v_0 = (150 {km}/h) = (150)(1000/3600) = 125/3 m/s = 41.67 m/s` .

Use (A) to write the stopping time in terms of *a*:

`0=41.67 m/s + a*t_"stopping"` , or

`t_"stopping"={-41.67}/a s`

Now, substitute the stopping time and stopping distance into (B) and solve for *a*:

`x-x_0=v_0*t_"stopping"+(a/2)*(t_"stopping")^2`

`50 m=(41.67 m/s)(-41.67/a)+(a/2)*({-41.67}/a)^2`

`50 m = {-41.67^2}/a+{41.67^2}/{2a}`

`a=-17.36 m * s^-2`

This corresponds to an acceleration of about 1.8 g.