Standard Form: Question 2

Worked solution

Part (a): Time for the probe

Use the relationship

$\text{time} = \frac{\text{distance}}{\text{speed}}.$

Substitute the values:

$t = \frac{7.2 \times 10^{11}}{1.6 \times 10^{4}}.$

Deal with the coefficients and the powers of ten separately.

Coefficients:

$\frac{7.2}{1.6} = 4.5.$

Powers of ten (subtract the indices when dividing):

$\frac{10^{11}}{10^{4}} = 10^{11-4} = 10^{7}.$

So

$t = 4.5 \times 10^{7}.$

The coefficient $4.5$ already satisfies $1 \le a < 10$, so this is in proper standard form.

$\boxed{t = 4.5 \times 10^{7}\ \text{s}}$

Part (b): Time for the radio signal

Again use $\text{time} = \dfrac{\text{distance}}{\text{speed}}$:

$t = \frac{7.2 \times 10^{11}}{3.0 \times 10^{8}}.$

Coefficients:

$\frac{7.2}{3.0} = 2.4.$

Powers of ten:

$\frac{10^{11}}{10^{8}} = 10^{11-8} = 10^{3}.$

So

$t = 2.4 \times 10^{3}.$

The coefficient $2.4$ is in the range $1 \le a < 10$, so no re-normalisation is needed.

$\boxed{t = 2.4 \times 10^{3}\ \text{s}}$

Part (c): How many times longer

We want the ratio

$\frac{\text{probe time}}{\text{signal time}} = \frac{4.5 \times 10^{7}}{2.4 \times 10^{3}}.$

Coefficients:

$\frac{4.5}{2.4} = 1.875.$

Powers of ten:

$\frac{10^{7}}{10^{3}} = 10^{7-3} = 10^{4}.$

So the ratio is

$1.875 \times 10^{4}.$

The coefficient $1.875$ satisfies $1 \le a < 10$, so this is already in proper standard form.

$\boxed{1.875 \times 10^{4}\ \text{times longer}}$

Alternative method for part (c)

You can avoid re-using rounded answers by dividing the original expressions:

$\frac{7.2 \times 10^{11} / 1.6 \times 10^{4}}{7.2 \times 10^{11} / 3.0 \times 10^{8}} = \frac{3.0 \times 10^{8}}{1.6 \times 10^{4}} = \frac{3.0}{1.6} \times 10^{8-4} = 1.875 \times 10^{4}.$

This confirms the result and shows that the ratio of times equals the ratio of the two speeds, $\dfrac{3.0 \times 10^{8}}{1.6 \times 10^{4}}$.

Check

A sanity check: $1.875 \times 10^{4} \approx 18\,750$. Multiplying the signal time by this factor, $2.4 \times 10^{3} \times 1.875 \times 10^{4} = 4.5 \times 10^{7}$ s, recovers the probe time in part (a). The answer is consistent.