The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration -5.40 m/s2 for 4.40 s, making straight skid marks 63.1 m long ending at the tree. With what speed does the car then strike the tree?

To determine the speed at which the car strikes the tree, we can use the equations of motion.

We know that the acceleration (a) is -5.40 m/s², the time (t) is 4.40 s, and the distance (s) is 63.1 m.

First, we can use the equation:
s = ut + (1/2)at²
where:
s = distance
u = initial velocity (which is what we're trying to find)
a = acceleration
t = time

Since the car starts from rest (u = 0), we can simplify the equation to:
s = (1/2)at²

Rearranging the equation to solve for acceleration gives us:
a = (2s) / t²

Substituting the given values, we have:
-5.40 m/s² = (2 * 63.1 m) / (4.40 s)²

Now, let's solve for the acceleration:
-5.40 m/s² = 9.0587 m / s² (after calculating)

Next, we can use the equation of motion:
v = u + at
where:
v = final velocity (which is what we're trying to find)
u = initial velocity (0 m/s)
a = acceleration (-5.40 m/s²)
t = time (4.40 s)

Rearranging the equation to solve for the final velocity gives us:
v = -at

Substituting the known values, we have:
v = -(-5.40 m/s²) * (4.40 s)

Calculating:
v = 23.76 m/s

Therefore, the car strikes the tree with a speed of 23.76 m/s.

jku