Question

If a particle has initial velocity of 3i +4j and an acceleration of 0.3i +0.4j it's speed after 10s is??????

Answer 1

v = u + at = 3i +4j + 10(0.3i +0.4j)
= 3i +4j +4i + 3j
= 7i +7j

so magnitude of final velocity is √(7²+7²)= 7√2

Answer 2

Answer:

7(2)^1/2

Explanation: given u=3i+4j

a=0.3i+0.4j

t=10s

u=v-at

3i+4j=v-(0.3i+0.4j)10

3i+4j=v-(3i+4j)

3i+4j+3i+4j=v

3+4(i+j)=v

7(i+j)=v

magnitude of v=root of 7^2+7^2

=root of 98

=7(2)^1/2