Question

The given table shows the distance traveled by a car in different time intervals.
$\begin{array}{cc}\text{Distance (in m)}& \text{Time (in s)}\\ 3.75& 2.5\\ 7.35& 3.5\\ 12.15& 4.5\end{array}$
Find the unit rate of change of speed of the car.

• A. $\frac{4}{5}{\text{m per s}}^2$
• B. $\frac{3}{5}{\text{m per s}}^2$
• C. $\frac{5}{3}{\text{m per s}}^2$
• D. $\frac{5}{4}{\text{m per s}}^2$

Answer 1

The correct option is B $\frac{3}{5}{\text{m per s}}^2$

Here, the unit rate of change in speed is asked.
We can find the speed of the car at every instant, as shown in the table.
$\begin{array}{ccc}\text{Distance, d (in m)}& \text{Time, t (in s)}& \text{Speed},s=\frac{d}{t}\text{(in m per s)}\\ 3.75& 2.5& \frac{3.75}{2.5}=\frac{3}{2}=1.5\\ 7.35& 3.5& \frac{7.35}{3.5}=\frac{735}{350}=\frac{147}{70}=2.1\\ 12.15& 4.5& \frac{12.15}{4.5}=\frac{1215}{450}=2.7\end{array}$
Let's find the ratio $\frac{s}{t}$.
$1.\frac{s}{t}=\frac{1.5}{2.5}=\frac{1.5\times 2}{2.5\times 2}=\frac{3}{5}{\text{m per s}}^2$
$2.\frac{s}{t}=\frac{2.1}{3.5}=\frac{2.1\times 10}{3.5\times 10}=\frac{21}{35}=\frac{3}{5}{\text{m per s}}^2$
$3.\frac{s}{t}=\frac{2.7}{4.5}=\frac{2.7\times 10}{4.5\times 10}=\frac{27}{45}=\frac{3}{5}{\text{m per s}}^2$

The ratio of the speed of the car and the time taken is constant.
Therefore, the speed of the car is proportional to the time taken.