## Spotting a police car, you brake your Porsche from a speed of 100 km/h to a speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration. a. What is that acceleration? b. How much time is required for the given decrease in speed?

Jul 2, 2018

a. $a = - 1.579 \frac{m}{s} ^ 2$
b. $t = 3.52 s$

#### Explanation:

We have velocity in km/hr and displacement in m. Since we have to make a unit conversion, might as well do it so it will yield the answer in the normal (SI) units for acceleration.

$100 \frac{k m}{h r} \cdot \frac{1000 m}{1 k m} \cdot \frac{1 h r}{3600 s} = 27.78 \frac{m}{s}$

$80.0 \frac{k m}{h r} \cdot \frac{1000 m}{1 k m} \cdot \frac{1 h r}{3600 s} = 22.22 \frac{m}{s}$

a. Using the formula of motion ${v}^{2} = {u}^{2} + 2 \cdot a \cdot d$

#(22.22 m/s)^2 = (27.78 m/s)^2 + 2a88.0 m

$493.7 {m}^{2} / {s}^{2} = 771.7 {m}^{2} / {s}^{2} + 176 m \cdot a$

$a = \frac{493.7 {m}^{2} / {s}^{2} - 771.7 {m}^{2} / {s}^{2}}{176 m} = - 1.579 \frac{m}{s} ^ 2$

b. Since the acceleration was constant, we can use the average velocity, and the distance to determine the time.

$t = \frac{88.0 m}{\frac{1}{2} \cdot \left(27.78 \frac{m}{s} + 22.22 \frac{m}{s}\right)} = 3.52 s$

As a double-check, if we convert $v = u + a \cdot t$ to $t = \frac{v - u}{a}$ and plug in the data, let's see if we get the same time.

$t = \frac{22.22 \frac{m}{s} - 27.78 \frac{m}{s} ^ 2}{- 1.579 \frac{m}{s} ^ 2} = 3.52 s$

I hope this helps,
Steve