To solve this question we will compute the magnitude of each vector.
Recall that the magnitude of a vector
[tex]a\hat{i}+b\hat{j}[/tex]
is:
[tex]\parallel a\hat{i}+b\hat{j}\parallel=\sqrt[]{a^2+b^2}.[/tex]
Therefore, the speed of the first car is:
[tex]\parallel26\hat{i}+31\hat{j}\parallel=\sqrt[]{26^2+31^2}\text{.}[/tex]
Simplifying the above equation we get:
[tex]\parallel26\hat{i}+31\hat{j}\parallel=\sqrt[]{676^{}+961}=\sqrt[]{1637}\approx40.46.[/tex]
The speed of the second car is:
[tex]\parallel40\hat{i}\parallel=\parallel40\hat{i}+0\hat{j}\parallel=\sqrt[]{40^2+0^2}\text{.}[/tex]
Simplifying the above equation we get:
[tex]\parallel40\hat{i}\parallel=\sqrt[]{40^2}=40\text{.}[/tex]
Answer:
The first car is the faster car.
Speed= 40.46.
