Question

Two cars $A$ and $B$ are at rest and at the same point initially. If $A$ starts with uniform velocity of $40\text{m/s}$ and $B$ starts in the same direction with constant acceleration of $4{\text{m/s}}^2$ , then $B$ will catch $A$ after the time

• A. $10\text{s}$
• B. $20\text{s}$
• C. $30\text{s}$
• D. $35\text{s}$

Answer 1

The correct option is B $20\text{s}$

Let $A$ and $B$ will meet after time $t\text{sec}$
As per the question the distance travelled by both will be equal.
Thus,
$S_A=ut=40t$ and $S_B=\frac{1}{2}a{t}^2=\frac{1}{2}\times 4\times t^2$
Since, $S_A=S_B\Rightarrow 40t=\frac{1}{2}4t^2\Rightarrow t=20\text{sec}$