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°=3 / 4 )

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°=3 / 4 ) (A) 8s (B) 12s (C) 16s (D) None

Tardigrade Admin · Accepted Answer

vA= tan 37=(3/4) vB= tan 53=(4/3) sA+4=SB (3/4) ⋅ t+4=(4/3)(t-4) 4+(16/3)=t((4/3)-(3/4)) (28/3)=t × (7/12) t=16

Q. 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 $3 7^{\circ} = 3/4$ )

8s

12s

16s

None

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

Q. 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∘=3/4 )

8s

12s

16s

None

vA​=tan37=43​ vB​=tan53=34​ sA​+4=SB​ 43​⋅t+4=34​(t−4) 4+316​=t(34​−43​) 328​=t×127​ t=16

Q. 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 $3 7^{\circ} = 3/4$ )

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