A ball is dropped from a height of 100 ft. one second later another ball is dropped from a height of 75 ft, which one hits the ground first? ...?

Answer: The final velocity is 33.78 mi/h, so the driver did reduced his speed enough.

Explanation: The kinetic energy of an object can be calculated as:

K = (1/2)m*v^2

We know that the mass of the car is m=1100kg

and the initial velocity is 22m/s

The initial kinetic energy is:

K = (1/2)*1100*(22)^2 = 266,200 joules.

Now, if the kinetic energy decreases by 1.4x10^5 J, the new kinetic energy is:

K = 266,200j - 140,000j = 126,200j

So we now can find the new velocity in m/s.

126,200 = (1/2)*1100*v^2

126,200*2/1100 = v^2

229.45 = v^2

v = (229.45)^(1/2) = 15.1 m/s

We know that the limit is 35 mi/h, so we need to transform our result into miles per hour.

We know that in one hour, there are 3600 seconds, so the velocity per hour is:

15.1*3600 m/h = 54,360 m/h

and we know that one mile is 1609.34 meters, so we need to divide by 1609.34.

v = (54,360/1609.34) mi/h = 33.78 mi/h

this is less than the speed limit, so the driver reduced his speed enough.

After losing kinetic energy, the car's final velocity was approximately 18.99 m/s, exceeding the exit's speed limit of 15.64 m/s, hence the driver did not reduce their speed adequately.

To determine whether the driver reduced their speed enough when exiting the highway, we must calculate the car's speed after its kinetic energy decreases by 1.4×105J. The initial kinetic energy (KE) of the 1100-kg car traveling at 22 m/s can be calculated using the equation KE = ½ mv². Plugging in the values, we can find the initial kinetic energy:

KEinitial = ½ (1100 kg)(22 m/s)² = 5.28×105J

After losing 1.4×105J of energy, the remaining kinetic energy will be:

KEfinal = KEinitial - 1.4×105J = (5.28 - 1.4)×105J = 3.88×105J

We can solve for the final velocity (vfinal) using the remaining kinetic energy:

½ (1100 kg)vfinal² = 3.88×105J

vfinal = √((2×3.88×105J) / 1100 kg)

vfinal ≈ 18.99 m/s

To compare to the speed limit, we convert 35 mi/h to meters per second:

35 mi/h × 0.44704 (conversion factor) = 15.64 m/s

Since the final velocity of the car is 18.99 m/s, which is greater than the exit's speed limit of 15.64 m/s, the driver did not reduce their speed enough.