Question

A particle has initial velocity (2i+3j​)m/s and has acceleration (0.4i+0.3j​)m/s2 . what is its speed after 10 minutes? ​

Answer 1

u = 3i + 4j

u = 3i + 4ja = 0.4i + 0.3j

u = 3i + 4ja = 0.4i + 0.3jv after 10s is:

u = 3i + 4ja = 0.4i + 0.3jv after 10s is:v =u +at

u = 3i + 4ja = 0.4i + 0.3jv after 10s is:v =u +at=7i +7j

thus, speed= 7√2

Answer 2

ANSWER

We have,

u =(3 i^+4 j^ )ms-¹

and

a=(0.4 i^+0.3 j^)ms -²

From I equation of motion after 10s

v = u + a t

=(3 i^ +4 j^ )+(0.4 i^ +0.3 j^)×10

=(3 i +4 j^ )+(4 i^+3 j^ )

⇒v=7 i^ +7 j^

⇒∣ v ∣=root 7 ²+7 ²

⇒∣ v ∣=7 root2 ms −¹