Q30P - Fundamentals Of Physics

Question

The brakes on your car can slow you at a rate of $\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a)If you are going  137 km/ h and suddenly see a state trooper, what is the minimum time in which you can get your car under the  90 km/ hspeed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun). (b)Graph x vs t and v vs t for such a slowing.

Step-by-Step Solution

Answer

a) Minimum time taken to get the car under  90 km/ h speed limit is 2.5 s.

Step 1: Given information

$$\begin{gathered} {\mathrm{v}}_0=137 \mathrm{km}/ \mathrm{h} \\ {\mathrm{v}}_{\mathrm{f}}=90 \mathrm{km}/\mathrm{h} \\ \mathrm{a}= -5.2 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Step 2: Concept and formula used in the given question

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Breaks of the car would generate deceleration, which would cause the car to slow down. The time required to bring the car under certain speed limits can be found by using kinematic equations which are given below.

Formula:

The final velocity in the kinematic equation is given by,

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}\times \mathrm{t}$                                                                                                                     (i)

Step 3: (a) Calculation for the minimum time in which you can get your car under the 90 km/ h speed limit

$$\begin{gathered} \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{time}, \mathrm{first} \mathrm{we} \mathrm{have} \mathrm{to} \mathrm{convert} \mathrm{the} \mathrm{speed} \mathrm{km}/\mathrm{h} \mathrm{into} \mathrm{m}/\mathrm{s}. \\ \frac{1 \mathrm{km}}{\mathrm{h}}=\frac{3600 \mathrm{s}}{1000 \mathrm{m}} \\ \frac{137 \mathrm{km}}{\mathrm{h}}\times \frac{1000 \mathrm{m}}{3600 \mathrm{s}}\times \frac{1 \mathrm{h}}{1 \mathrm{km}}=38.1 \mathrm{m}/\mathrm{s} \\ \frac{90 \mathrm{km}}{\mathrm{h}}\times \frac{1000 \mathrm{m}}{3600\mathrm{s}}\times \frac{1 \mathrm{h}}{1 \mathrm{km}}=25 \mathrm{m}/\mathrm{s} \\ \mathrm{Using} \mathrm{equation} \left(\mathrm{i}\right), \\ 25=38.1+(-5.2)\times \mathrm{t} \\ \mathrm{So}, \\ t=2.5 s \end{gathered}$$

Step 4: (b) Plotting the graph x vs t and v vs t for such a slowing

Graph of x(t) vs t is as shown below: