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Q4.7-50PE

Expert-verifiedFound in: Page 164

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(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/s ^{2} for 50.0 s? **

**(b) What is unreasonable about the result? **

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

(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/s^{2} in 50 seconds.

- The velocity of the car = 50 km/h.
- Car decelearate at a rate of .400 m/s
^{2 }for 50 seconds.

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:

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

Substitute 13.89 m/s for *u*, (-0.4) m/s^{2} 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**.

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.

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/s^{2} in 50 seconds.

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