Q10P - Fundamentals Of Physics

Question

To set a speed record in a measured (straight line) distance d, a race car must be driven in one direction (in time $t_1$ ) and then in the opposite direction (in time $t_2$ ). (a) To eliminate the effects of wind and obtain the car’s speed in a windless situation, should we find the average of $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$$}\right.t_1$ and $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$t_2$}\right.$ (method 1) or should we divide d  by the average of $t_1$ and $t_2$? (b) What is the fractional difference in the two methods when a steady wind blows along the car’s route and the ratio of wind speed $V_w$  to car’s speed $V_c$  is 0.0240.

Step-by-Step Solution

Answer

1. We should find the average of $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$t_1$}\right.$ and $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$t_2$}\right.$ to eliminate the effect of wind in obtaining the car’s speed.
2. Fractional difference in the two methods is $5.76\times {10}^{-4}$. 

Step 1: Given data

The ratio of wind speed $V_w$  to car’s speed $V_c$ is 0.0240.

Step 2: Understanding the velocity of an object

Velocity may be defined as the time rate of change of displacement. It is a vector quantity. The velocity of the wind would be added to the velocity of the car when both are moving in the same direction. When a car and wind are in opposite directions, velocity would be the difference between them. 

The expression for the velocity is given as follows:

$v=\frac{\Delta x}{\Delta t}$                                           .......(i)

Here,  $\Delta X$ is the displacement, and $\Delta t$ is the time duration. 

Step 3: (a) Determination of the speed using the first method

Let $V_w$  be the speed of the wind, and $V_c$  be the speed of the car.

Suppose, during time interval $t_1$ , the car moves in the same direction as the wind. Then the effective speed of the car is calculated as follows:

$v_{eff1}=v_c+v_w$

Then the distance traveled is calculated as follows:

$d= v_{eff1}\times t_1$

On the other hand, for the return trip during a time interval $t_2$, the car moves in the opposite direction of the wind, and the effective speed will be calculated as follows

$v_{\mathrm{eff}2}=v_c-v_w$

And the distance traveled is calculated as follows:

$d=v_{eff2}\times t_2$

The above two equations can be written as follows:

$v_c+v_w=\frac{d}{t_1}$

$v_c-v_w=\frac{d}{t_2}$

By adding these two equations, you will get

$v_c=\frac{1}{2}\times \left(\frac{d}{t_1}+\frac{d}{t_2}\right)$

Method 1 eliminates the effect of wind in obtaining the car’s speed which was calculated to be

$v_c=\frac{1}{2}\times \left(\frac{d}{t_1}+\frac{d}{t_2}\right)$

Step 4: Determination of the speed using the second method

Using method 2,

$\begin{array}{c}{v\text{'}}_c=\frac{d}{\frac{t_1+t_2}{2}}\\ =\frac{2d}{t_1+t_2}\end{array}$

Putting the values of  $t_1$and  $t_2$,