Question

The car will take approximately:

a) 5.75 seconds
b) 6.00 seconds
c) 5.50 seconds
d) 6.25 seconds

Answer 1

Final answer:

The question is about calculating the speed at which a 6.0 m-long sports car must move to appear 5.5 m long due to Lorentz contraction, a concept in special relativity within the field of Physics, commonly studied in High School.

Step-by-step explanation:

Understanding the Physics Question

The question you are asking is related to the concept of special relativity, specifically the Lorentz contraction effect, which is a physics concept that falls under the subject area of Physics and is typically covered at the High School level or above. The question is asking for the speed at which a 6.0 m-long sports car must travel so that it appears to be only 5.5 m long to an observer. This phenomenon can be explained by the equation for Lorentz contraction:

L' = L / γ, where γ (gamma) is the Lorentz factor, calculated as 1/√(1 - v²/c²), with v being the velocity of the car and c being the speed of light.

To solve for v, we need to rearrange the Lorentz contraction formula to isolate v, and calculate it considering that the original length L is 6.0 m and the observed contracted length L' is 5.5 m. This will involve some algebraic manipulation and substitution of values into the formula to find the speed v that results in the specified contraction.

Answer 2

In this case, The car will take approximately: 3.33 seconds to stop

The answer is option ⇒e

Step-by-step explanation:

To determine the time it takes for the car to stop, we can use the kinematic equation:

v² = u² + 2as

Where:

• v = final velocity (0 m/s, as the car comes to a stop)
• u = initial velocity (10.0 m/s, given in the question)
• a = acceleration (-3.0 m/s², given in the question)
• s = distance covered

Rearranging the equation to solve for s:

s = (v² - u²) / (2a)

Substituting the given values:

s = (0² - 10.0²) / (2(-3.0))

s = (0 - 100.0) / (-6.0)

s = 16.67 meters (rounded to two decimal places)

Now that we know the distance covered (s), we can calculate the time it takes for the car to stop using the formula:

t = (v - u) / a

Substituting the given values:

t = (0 - 10.0) / (-3.0)

t = 3.33 seconds (rounded to two decimal places)

The answer is option ⇒e)3.33 seconds

Your question is incomplete, but most probably the full question was:

A braking car on dry pavement has a constant acceleration of -3.0 m/s². If the car was moving at 10.0 m/s

The car will take approximately:_ to stop

a) 5.75 seconds

b) 6.00 seconds

c) 5.50 seconds

d) 6.25 seconds

e)3.33 seconds