Question

An object has velocity $\overrightarrow{u}=\hat{i}+\hat{j}m/s$ at time $t=0$ It undergoes a constant acceleration$a=\left(\hat{i}-2\hat{j}\right)m/s^2$ for $4sec$ The magnitude of velocity at the end of $4sec$ is

• A. $\sqrt{74}m/s$
• B. $\sqrt{80}m/s$
• C. $\sqrt{97}m/s$
• D. $\sqrt{47}m/s$

Answer 1

The correct option is A

$\sqrt{74}m/s$

The explanation for the correct option

Step 1: Given

Initial velocity $\overrightarrow{u}=\hat{i}+\hat{j}m/s$

Acceleration $a=\left(\hat{i}-2\hat{j}\right)m/s^2$

Step 2: Formula used

$\stackrel{\rightharpoonup }{v}=\overrightarrow{u}+\overrightarrow{a}t$

Where $\overrightarrow{u}$ is the initial velocity, $\overrightarrow{v}$ is the final velocity $\overrightarrow{a}$ is the acceleration $t$ is the time

Step 3: Find the velocity

$$\begin{gathered} \overrightarrow{u}=\hat{i}+\hat{j}m/s \\ t=4sec \end{gathered}$$

$a=\left(\hat{i}-2\hat{j}\right)m/s^2$

$$\begin{gathered} \therefore \stackrel{\rightharpoonup }{v}=\overrightarrow{u}+\overrightarrow{a}t \\ \Rightarrow \stackrel{\rightharpoonup }{v}=\hat{i}+\hat{j}+4\left(\hat{i}-2\hat{j}\right) \\ =5\hat{i}-7\hat{j} \\ \left|\overrightarrow{v}\right|=\sqrt{5^2+7^2} \\ =\sqrt{74}m/s \end{gathered}$$

Hence option A is the correct answer.