Q4.7-50PE - College Physics (Urone)

Question

(a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of 0.400 m/s² for 50.0 s? 

(b) What is unreasonable about the result?

(c) Which premise is unreasonable, or which premises are inconsistent?

Step-by-Step Solution

Answer

(a) The final velocity is (-6.11) m/s.

(b) It is unreasonable that on the application of brakes, the car will move in the opposite direction.

(c) The unreasonable premise is that the car traveling at a speed of 13.89 m/s can decelerate at a rate of 0.4 m/s² in 50 seconds.

Step 1: Given Data

• The velocity of the car = 50 km/h.
• Car decelearate at a rate of .400 m/s² for 50 seconds.

Step 2: (a) Determine the final velocity of the car.

Apply the following equation of motion as:

 

$\mathit{\text{v}}\mathit{=}\mathit{\text{u}}\mathit{+}\mathit{\text{at}}$

 

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Convert the initial speed of the car from 50.0 km/h to m/s as:

 

$50\text{\hspace{0.33em}km/h}=50\frac{\text{km}}{\text{h}}\times \frac{1000\text{\hspace{0.33em}m}}{1\text{\hspace{0.33em}km}}\times \frac{1\text{\hspace{0.33em}h}}{3600\text{\hspace{0.33em}s}}=13.89\text{\hspace{0.33em}m/s}$

 

Substitute 13.89 m/s for u, (-0.4) m/s² for a, and 50 s for t in the above expression, and we get,

 

$\begin{array}{c}v=13.89\text{\hspace{0.33em}m/s}-0.4{\text{\hspace{0.33em}m/s}}^{\text{2}}\times 50\text{\hspace{0.33em}s}\\ =13.89\text{\hspace{0.33em}m/s}-20\text{\hspace{0.33em}m/s}\\ =-6.11\text{\hspace{0.33em}m/s}\end{array}$

 

Hence, the final velocity is -6.11 m/s.

Step 3: (b) the unreasonability of the result

The result shows that the car moves in the opposite direction. It is unreasonable that on the application of brakes, the car will move in the opposite direction.

Step 4: (c) the premises that are unreasonable

The unreasonable premise is that the car traveling at a speed of 13.89 m/s can decelerate at a rate of 0.4 m/s² in 50 seconds.