Question

On a two-lane road, car A is travelling with a speed of $v=5{\text{ms}}^{-1}$. Two cars, B and C approach car A in opposite directions with a speed of $u=10{\text{ms}}^{-1}$ each. At a certain instant when the B and C are equidistant from A each being $l=1500\text{m}$, B decides to overtake A before C passes A. What minimum acceleration(in ${\text{m/s}}^2$) of car B is required to avoid an accident with C ?

Answer 1

In a frame attached to A,
Velocity of B relative to A,
$v_{BA}=5\text{m/s}$
and velocity of C relative to A,
$v_{CA}=15\text{m/s}$
For minimum acceration, let B and C pass A at same instant.
Hence both cover the distance $l$ in time t
For car B, $l=5t+\frac{1}{2}a{t}^2$
For car C, $l=15t$
As it is moving with constant speed
$\Rightarrow t=\frac{l}{15}=\frac{1500}{15}=100s$
Substituting values of l and t in equation (1), we get
$1500=5\left(100\right)+\frac{1}{2}a(100{)}^2$
$1000=\frac{a}{2}\left(10000\right)$
$a=\frac{2}{10}=0.2{\text{m/s}}^2$