A car is speeding along the motorway at 40 m s when the driver

observes an accident ahead. He slams on the brakes, the wheels lock immediately, and the car skids to a halt. You may assume that the car travels in a straight line whilst it is skidding and that the magnitude of its deceleration is 10 m s^-2.
What distance does the car travel between the time that the brakes are applied and the time that the car comes to rest, and how much time does this take?

time to stop = t

v = Vi - 10 t
0 = 40 - 10 t
t = 4 seconds to stop
d = average speed during stop * time = 20 * 4 = 80 meters

a. V^2 = Vo^2 + 2a*d = 0,

40^2 + (-20)d = 0,
d = 80 m.

b. V = Vo + a*t = 0,
40 + (-10t = 0,
t = 4 s.

To find the distance the car travels while braking and the time it takes to come to a halt, we can use equations of motion.

First, we need to find the time it takes for the car to come to rest. We can use the following equation:

v = u + at

Where:
v = final velocity (0 m/s, since the car comes to rest)
u = initial velocity (40 m/s)
a = acceleration (-10 m/s²)

0 = 40 + (-10)t

Simplifying the equation, we have:

0 = 40 - 10t

Solving for t, we can move 40 to the other side of the equation:

10t = 40

t = 40/10
t = 4 seconds

So, it takes 4 seconds for the car to come to a halt.

Next, we can find the distance traveled using the equation:

s = ut + (1/2)at²

s = (40)(4) + (1/2)(-10)(4²)

Simplifying the equation:

s = 160 - 80

s = 80 meters

Therefore, the car travels a distance of 80 meters between the time the brakes are applied and the time it comes to rest, and it takes 4 seconds to do so.