Question

Two cars, initially at a separation of $12\text{m}$, start simultaneously. First car $A$, starting from rest moves with an acceleration $2{\text{m/s}}^2$ , whereas the car $B$, which is ahead, moves with a constant velocity $1\text{m/s}$, away from car $A$ along the same direction. Find the time when car $A$ overtakes car $B$.

• A. $4\text{s}$
• B. $6\text{s}$
• C. $5\text{s}$
• D. $7\text{s}$

Answer 1

The correct option is A $4\text{s}$

Relative velocity of $A$ with respect to $B$ is $v_{AB}$,

At the start of the motion

$v_{AB}=v_A-v_B=0-1=-1\text{m/s}$

Relative acceleration of $A$ with respect to $B$ is
$a_{AB}=a_A-a_B=2-0=2{\text{m/s}}^2$

Applying equation of motion for a single body (remember the second body is rendered stationary).

$x=ut+\frac{1}{2}a{t}^2$ where $(x=12\text{m})$, initial relative velocity $(v_{AB}=1\text{m/s})$ and $a_{AB}=2{\text{m/s}}^2$, we get

$12=-1t+\frac{1}{2}\times 2\times t^2$
$\Rightarrow t^2-t-12=0$,
by solving for $t$ we get
$t=4\text{s},t=-3\text{s}$
$\Rightarrow t=4\text{s}$ As, time cannot be negative,