A car travels at a constant speed of 34.5 m/s for a duration of 15...

The distance covered is given by the product of speed and time.

$d = v \times t$

$d = 34.5 \times 15 = 517.5\text{ m}$

The number of significant figures in the speed ($34.5\text{ m/s}$) is $3$.

The number of significant figures in the time ($15\text{ s}$) is $2$.

According to the rules of significant figures for multiplication, the final result must be reported to the same number of significant figures as the measurement with the least number of significant figures. Therefore, the result must be rounded to $2$ significant figures.

Rounding $517.5$ to $2$ significant figures gives $520\text{ m}$, which is expressed in scientific notation as $5.2 \times 10^{2}\text{ m}$.

Answer: $5.2 \times 10^{2}\text{ m}$