*Please solve the solutions step by step with correct values, thanks in advance!
ID: 2173394 • Letter: #
Question
*Please solve the solutions step by step with correct values, thanks in advance!
A particle of mass m = 0.1 kg has a velocity vector, v, given by; v- = 3.2i- + 4.3 j- + 5.4 k- m s-1 it is acted on by a force f- = 7.1 i- + 8.2 j- + 9.3 k- N Find the speed of the particle. Write down a unit vector in the direction of velocity. Find the angle between the projection of the velocity vector in the x, y plane and the x-axis. Find an expression for the acceleration. Find the velocity at a time 2 s later.Explanation / Answer
speed=magnitude of velocity vector=sqrt(3.2^2+4.3^2+5.4^2)=7.6058 m/s (ii.) unit vector along velocity=(3.2 i +4.3 j +5.4 k)/7.6058 (iii) projection on x,y plane=3.2i+4.3 j projection on x-axis=3.2 i so angle=53.343 degrees (iv.) acceleration=force/mass=71i +82 j +93 k v. velocity after t=2 v+ 2*a=145.2 i +168.3 j+ 191.4 k
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