Question

The position time graphs for two particles $A$ and B moving in same straight line are shown in figure. The time after which (from when A starts) B caught A. (tan $37^{\circ}=3 / 4$ )

• A. 8s
• B. 12s
• C. 16s
• D. None

Answer 1

Answer: (C)

$v_{A}=\tan 37=\frac{3}{4}$
$v_{B}=\tan 53=\frac{4}{3}$
$s_{A}+4=S_{B}$
$\frac{3}{4} \cdot t+4=\frac{4}{3}(t-4)$
$4+\frac{16}{3}=t\left(\frac{4}{3}-\frac{3}{4}\right)$
$\frac{28}{3}=t \times \frac{7}{12}$
$t=16$