Question

A particle is located at position $(3\hat{i}+4\hat{j})m$ and starts from rest under acceleration $(2\hat{i}-\hat{j})m{s}^{-2}$ then at what time its position will be $(19\hat{i}-4\hat{j})m$?

• A. $4s$
• B. $2s$
• C. $\frac{1}{2}s$
• D. $5s$

Answer 1

The correct option is A $4s$

using : $\overrightarrow{s}=\overrightarrow{u}t+\frac{\overrightarrow{a}{t}^2}{2}$

$(19\hat{i}-4\hat{j})-(3\hat{i}+4\hat{j})=\frac{(2\hat{i}-\hat{j})}{2}{t}^2$

$32\hat{i}-16\hat{j}=(2\hat{i}-\hat{j})t^2$

By compairing components

$32=2t^2$$\cap$$16=t^2$

$t=4s$