Question

If a particle is kept at rest at origin, another particle starts from $(5,0)$ with a velocity of $-4\hat{i}+3\hat{j}$ . Find their closest distance of approach.

• A. $3m$
• B. $4m$
• C. $5m$
• D. $2m$

Answer 1

The correct option is A $3m$

The position of the particle at the time $t$ is given as,

$D=\left(5-4t\right)\hat{i}+3t\hat{j}$

The distance from the particle at the origin is given as,

$D=\sqrt{{\left(5-4t\right)}^2+3t^2}$ (1)

For displacement to be minimum $\frac{dD}{dt}$ have to be zero, it can be written as,

$\frac{dD}{dt}=0$

$\frac{50t-40}{2\sqrt{{\left(5-4t\right)}^2+3t^2}}=0$

$t=\frac{4}{5}s$

By substituting the value of $t$ in the equation (1), we get

$D=\sqrt{{\left(5-4\times \frac{4}{5}\right)}^2+{\left(3\times \frac{4}{5}\right)}^2}$

$D=3m$

Thus, the closest distance of approach is $3m$.