Question

A particle has initial velocity (2i+3j) and acceleration (0.3i+0.2j) the magnitude of the velocity after 10 second will be

Answer 1

Here is it answer...

Answer 2

Answer:

The magnitude of velocity after 10 seconds is $\bold{{|v|=5 \sqrt{2} m / s}}$

Explanation:

The particle moves with initial velocity and acceleration as given, the velocity is calculated using equation of motion

Given is the initial velocity u = 2 i + 3 j  

Acceleration a = 0.3 i + 0.2 j

Final velocity v =?

Time t = 10sec

From first equation of motion, we have –

v=u+a $\bold{\times}$ t

v = u + at

v=(2 i+3 j)+(0.3 i+0.2 j) $\times$10

v = 2 i + 3 j + 3 i + 2 j

v = 5 i + 5 j

$\begin{array}{l}{|v|=|5 i+5 j|} \\ {|v|=\sqrt{5^{2}+5^{2}}=\sqrt{25+25}=\sqrt{100}}\end{array}$

$\begin{array}{l}{|v|=\sqrt{5 \times 5 \times 2}} \\ \bold{{|v|=5 \sqrt{2} m / s}\end{array}}$