Question

On a dry road, a car with good tires may be able to brake with a constant deceleration of 5.74 m/s^2. (a) How long does such a car, initially traveling at 29.3 m/s, take to stop? (b) How far does it travel in this time ?

Answer 1

Time needed to stop: `"5.10 s"`
Distance covered: `"74.8 m"`

Explanation:

This time, you know that at `t=0` the car has a speed of `"29.3 m/s"`.

At this moment, it starts slowing down with a constant deceleration of `"5.74 m/s"""^2`. After covering a distance `d`, the car comes to a complete stop.

This means that you can write

`underbrace(v^2)_(color(blue)(=0)) = v_0^2 - 2 * a * d`

The negative sign is there to show that the acceleration is oriented in the opposite direction to the direction of movement, which is what you would expect for a car that's slowing down.

So, you have

`d = v_0^2/(2 * a) = (29.3""^2"m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 5.74color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = color(green)("74.8 m")`

The time needed for the car to come to a complete stop is

`underbrace(v)_(color(blue)(=0)) = v_0 - a * t`

`t = v_0/a = (29.3color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(5.74color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = color(green)("5.10 m/s")`