Q28P - Fundamentals Of Physics

Question

On a dry road, a car with good tires may be able to brake with a constant deceleration of $\mathbf{4}\mathbf{.}\mathbf{29}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$ . (a)How long does such a car, initially travelling at 24.6 m/s, take to stop? (b)How far does it travel in this time? (c)Graph x vs t and v vs t for the deceleration.

Step-by-Step Solution

Answer

1. Time is 5.00 s
2. Distance is 61.5 m 

Step 1: Given information

$$\begin{gathered} {\mathrm{v}}_0=24.6 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}=0 \mathrm{m}/\mathrm{s} \\ \mathrm{a}=-4.92 \mathrm{m}/\mathrm{s} \mathrm{Here} \mathrm{negative} \mathrm{sign} \mathrm{because} \mathrm{of} \mathrm{deceleration}. \end{gathered}$$

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. Kinematic equations can be used to find the time taken by the car to stop. Another kinematic equation would give the distance traveled by the car in that time. Further, the displacement and velocity for different time intervalscan be found to plot the graph.

Formula:

The final velocity in kinematic equations can be expressed as,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}\times \mathrm{t} \left(1\right) \\ {{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{a}\times \mathrm{x} \\ \left(2\right) \\ \mathrm{Where} {\mathrm{v}}_0 \mathrm{is} \mathrm{the} \mathrm{initial} \mathrm{velocity} \mathrm{a} \mathrm{is} \mathrm{an} \mathrm{acceleration} \mathrm{t} \mathrm{is} \mathrm{the} \\ \mathrm{time} \mathrm{and} \mathrm{x} \mathrm{is} \mathrm{the} \mathrm{displacement}. \end{gathered}$$

Step 3: (a) Calculation for the how long does such a car, initially travelling at 24.6 m/s , take to stop

$$\begin{gathered} \mathrm{Using} \mathrm{equation} \left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}\times \mathrm{t} \\ 0=24.6+(-4.92)\times \mathrm{t} \\ \mathrm{t}=5.0 \mathrm{s} \end{gathered}$$

Step 4: (b) Calculation for the how far does it travel in this time

$$\begin{gathered} \mathrm{Using} \mathrm{eq}+\mathrm{uation} \left(\mathrm{ii}\right), \\ {{\mathrm{v}}_{\mathrm{f}}}^2={{\mathrm{v}}_0}^2+2\mathrm{a}\times \mathrm{x} \\ \mathrm{x}=\frac{{{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}_0}^2}{2\times \mathrm{a}} \\ =\frac{0^2-24.6^2}{2\times (-4.92)} \\ =61.5 \mathrm{m} \end{gathered}$$

Step 5: (c) Plotting the graph x vs t and v vs t for the deceleration