Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Ship $A$ is moving with velocity $V_{1}=30 i+50 i$ from position (0,0) and ship $B$ is moving with velocity $V_{2}=-10 i$ from position (80,150). The time for minimum separation between the two ships is:(A) 2.6(B) 2.2(C) 2.4(D) None

Last updated date: 20th Sep 2024
Total views: 81.6k
Views today: 1.81k
Verified
81.6k+ views
Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

$\overrightarrow{\mathrm{V}}_{\mathrm{r}}=40 \hat{\mathrm{i}}+50 \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{r}}_{\mathrm{r}}=-80 \hat{\mathrm{i}}-150 \hat{\mathrm{j}}$
$t_{min}$= ${|{\dfrac{{\overrightarrow V}_{r} \cdot {\overrightarrow r}_{r}}{\overrightarrow V}}|}^2$ = $\dfrac{10700}{4100} = \dfrac{107}{41} = 2.6s$