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 ?

1 Answer
Sep 9, 2015

Answer:

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")#