1. Motion III Acceleration is not constant

2. Up until now, we have only considered motion where the acceleration is constant with respect to time. • Now let us consider the general case of time dependent acceleration.

3. Forces on a Body • Consider a body of mass m subject to many time dependent force vectors. • Taking the vector sum of all the force vectors applied to the body, we get • Σ F = (t + 1)i + (t2 + 3t – 4)j + (2)k • a = [(t + 1)/m]i+ [(t2 + 3t – 4)/m]j +(2/m)k • Motion in each direction is independent of motion in the other directions, so

4. ax= [(t + 1)/m]i • ay= [(t2 + 3t – 4)/m]j • az= (2/m)k • Remembering that ∫dv = ∫adt • We can integrate the left side from voto v • And the right side from toto t. • For the x, the y, and the z directions, but a must be kept inside the integral because it is time dependent!

5. vx= vox + ∫[(t + 1)/m]idt • vx= vox+ (1/m)[(t2/2) + t]i • vy= voy+ ∫[( t2 + 3t – 4)/m]jdt • vy= voy+ (1/m)[(t3/3) + (3t2/2) – 4t]j • vz= voz + ∫[(-4)/m]kdt • vz= voz - (4t)/m]k

6. The three dimensional velocity vector equation is now: • v = vx + vy + vz • v = vox+ (1/m)[(t2/2) + t]i+ voy+ (1/m)[(t3/3) + (3t2/2) – 4t]j + voz- (4t)/m]k

7. To obtain the position vectors x, y, z, one must integrate again • ∫dx = ∫vxdt • x = xo+ ∫{vox+ (1/m)[(t2/2) + t]i}dt • x = xo+ voxt + (1/m)[(3t3/6) + (t2/2]i • ∫dy = ∫vydt • y = yo+ ∫{voy+ (1/m) )[(t3/3) + (3t2/2) – 4t]j}dt • y = yo+ voyt + (1/m) )[(t4/12) + (t3/2) – 2t2

8. ∫dz = ∫vzdt • z = zo+ ∫{voz+ (1/m)[(-4 t]k}dt • z = zo+ vozt + (1/m)[(-2 t2]k

9. The final position vector r is a function of time and would be the vector sum • r = x + y + z • r = xo+ voxt + (1/m)[(3t3/6) + (t2/2]i+ yo+ voyt + (1/m) )[(t4/12) + (t3/2) – 2t2]j + zo+ vozt + (1/m)[(-2 t2]k