When exiting the highway, a 1100-kg car is traveling at 22 m/s. The car's kinetic energy decreases by 1.4×105J The exit's speed limit is 35 mi/h. Did the driver reduce its speed enough?

Answers

Answer 1

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.

Answer 2
Final answer:

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.

Explanation:

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.


Related Questions

If a force of 10 n is applied to an object with a mass of 1kg the object will accelerate at

Answers

We Know, F = m*a
Here, F = 10 N
m = 1 Kg

Substitute their values in the equation,
10 = 1 * a
a = 10/1
a = 10

So, your final answer & the acceleration of the object would be 10 m/s²

Hope this helps!
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