Question

A particle initial velocity of 3 ICAP + 4 j cap and an acceleration of 0.4 ICAP + 0.6 J cap it speed after 10 second is​

Answer 1

Answer:

$7i + 10j$

Explanation:

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REFER ATTACHMENT.

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Answer 2

Explanation:

initial velocity =u=>3i+4j

acceleration=a=>0.4i+0.6j

v=u+at

Final velocity after 10 seconds =>

v=(3i+4j)+10(0.4i+0.6j) =>v= 3i+4j+4i+6j =7i+10j So it's speed after 10 seconds =>

$\sqrt{ 7^{2} + {10}^{2}} = \sqrt{49 + 100} = \sqrt{149} =$

12.2 approximately. Hope it helps you.