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 speedin 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.

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

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.

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

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

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 $t_2$,