Question

A particle has initial velocity ( 3 i^{^} + 4 j^{^} ) and has acceleration ( 0.4 i^{^} + 0.3 j^{^} ). It's speed after 10 seconds will be?
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Answer 1

Given
Vi= ( 3 i^{^} + 4 j^{^} )
a= ( 0.4 i^{^} + 0.3 j^{^} ).
t=10s
Using the kinematic equation; Vf = Vi+at = ( 3 i^{^} + 4 j^{^} ) +( 0.4 i^{^} + 0.3 j^{^} ).10 = ( 3 i^{^} + 4 j^{^} ) + ( 4 i^{^} + 3 j^{^} ) = ( 7i^{^} + 7 j^{^} )
In vector notation we have Vf = ( 7i^{^} + 7 j^{^} ) and magnitude as Vf = [ 7^2+7^2] ^ 1/2= 9.99 m/s
Direction- tan θ= Vfy / Vfx= 7/7= 1
θ= tan^ 1[1]= 45 °

Answer
Vf = ( 7i^{^} + 7 j^{^} )
Magnitude = 9.99 m/s~10 m/s
Direction= 45 °