Question

a particle has initial velocity 3i^+4j^and an acceleration of 0.3i^+0.4j^ find its displacement after 10 seconds​

Answer 1

Answer:

Explanation:

the initial velocity is give as vi = 3i + 4j  

the acceleration is a = 0.4i + 0.3j  

and the time elapsed is t = 10s

now final velocity after 10s is vf = (3i+4j)m/s + 10s(0.4i+0.3j)m/s

vf = 7i + 7j

now speed is v =\sqrt{(7i)^2)+ (7j))} = 7\sqrt{2}

the initial velocity is give as vi = 3i + 4j  

the acceleration is a = 0.4i + 0.3j  

and the time elapsed is t = 10s

now final velocity after 10s is vf = (3i+4j)m/s + 10s(0.4i+0.3j)m/s

vf = 7i + 7j

now speed is v =\sqrt{(7i)^2)+ (7j))} = 7\sqrt{2}

Answer 2

Answer:75

Explanation: