Question

In a bicycle competition called Tour De France, two riders; $A$ and $B$ are riding bicycles at constant acceleration of magnitude $1{\text{m/s}}^2$ and $3{\text{m/s}}^2$ respectively, in the North direction. Their initial speeds are $15\text{m/s}$ and $10\text{m/s}$ respectively while they were heading in the North direction. Find the separation between them after $t=10\text{s}$.

• A. $80\text{m}$
• B. $95\text{m}$
• C. $50\text{m}$
• D. $125\text{m}$

Answer 1

The correct option is C $50\text{m}$

Let the north direction is represented by $\hat{i}$
Initial velocity of $A,{\overrightarrow{u}}_A=15\hat{i}\text{m/s}$
Initial velocity of $B,{\overrightarrow{u}}_B=10\hat{i}\text{m/s}$
Initial velocity of $A$ w.r.t. $B$, ${\overrightarrow{u}}_{AB}={\overrightarrow{u}}_A-{\overrightarrow{u}}_B$
$=15\hat{i}-10\hat{i}=5\hat{i}\text{m/s}$
Acceleration of $A,{\overrightarrow{a}}_A=\hat{i}{\text{m/s}}^2$
Acceleration of $B,{\overrightarrow{a}}_A=3\hat{i}{\text{m/s}}^2$
Acceleration of $A$ w.r.t. $B$, ${\overrightarrow{a}}_{AB}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B$
$=\hat{i}-3\hat{i}=-2\hat{i}{\text{m/s}}^2$
So, separation between $A$ and $B$ after $t=10$ can be given by
$\therefore {\overrightarrow{x}}_{AB}={\overrightarrow{u}}_{AB}t+\frac{1}{2}{\overrightarrow{a}}_{AB}{t}^2$ (using second equation of motion)
$=5\hat{i}\times 10+\frac{1}{2}(-2\hat{i})\times {10}^2$
$=(50-100)\hat{i}=-50\hat{i}\text{m}$
$\therefore$ After time $t=10$ separation between them will be $50\text{m}$