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 $1\text{m/s}$ along the same direction. Find the time when the car "$A$" overtakes car "$B$".

• A. $4\text{s}$
• B. $7\text{s}$
• C. $6\text{s}$
• D. $8\text{s}$

Answer 1

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

$\text{I}-$ Method

Relative velocity of "$A$" with respect to "$B$", i.e $v_{AB}$, at the start of motion is :
$v_{AB}=v_A-v_B=0-1=1\text{m/s}$
Relative acceleration of "$A$" with respect to "$B$". i.e. $a_{AB}$ 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).
$x=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 12=-1\times t+\frac{1}{2}\times 2\times t^2$
$\Rightarrow t^2-t-12=0$
$\Rightarrow t^2+3t-4t-12=0$
$\Rightarrow (t+3)(t-4)=0$
$\Rightarrow t=4\text{s}$

$\text{II}-$ Method
The car $A$ will cover $12\text{m}$ more than car $B$

Hence, displacement of $A=$ displacement of $B+12$

Now equating the displacement by $2\text{nd}$ equation of motion

$\frac{1}{2}a{t}^2=vt+12$
$\Rightarrow \frac{1}{2}\left(2\right)t^2=1\times t+12$

$\Rightarrow t^2-t-12=0$
$\Rightarrow (t+3)(t-4)=0$
$\Rightarrow t=4\text{s}$