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Kinematics of a Particle in a Plane (examples of rectangular description)

Example: The velocity of particle P moving in the x-y plane is given by

v_P = t [ sin πt/4 i + cos πt/4 j ] ft/sec

Find the acceleration of the particle at time t = 2 seconds.

Solution: The acceleration of a particle is the time rate of change of the velocity with

respect to the frame of reference.

Strategy: Take the derivative of the velocity with respect to time and evaluate at t = 2 sec.

a_P = d v_P /dt = [ sin πt/4 i + cos πt/4 j ] + t [ π/4 cos πt/4 – π/4 sin πt/4 ]

at t = 2 sec a_P = sin π/2 i + cos π/2 j + 2 (π/4) [ cos π/2 i - sin π/2 j ]

therefore the result is: a_P = i - π/2 j ft/sec²

Example: The displacement over time is the difference in the position vectors over the same time.

Suppose the velocity of the particle is the same as in the first example above. Find the

displacement of the particle over the interval from t = 4 seconds to t = 6 seconds.

Solution: Integrate the velocity to find the position vector. Then evaluate the position

vector from t = 4 seconds to t = 6 seconds.

t = 6

r_P(t) = ∫ t [ sin πt/4 i + cos πt/4 j ] dt Strategy: use integration by parts

t = 4

u = t dw = [ sin πt/4 i + cos πt/4 j

du = dt w = -4/π cos πt/4 i + 4/π sin πt/4 j

6 6

r_P(6) - r_P(4) = t [-4/π cos πt/4 i + 4/π sin πt/4 j | 16/π² sin πt/4 i - 16/π² cos πt/4 i |

4 4

Displacement = 24/π [- cos 3π/2 i – sin 3π/2 j] + 16/π² [sin 3π/2 i – cos 3π/2 j]

- 16/π [ - cosπ i + sinπ j ] + 16/π² [ sinπ i – cosπ j ]

Displacement = (16/π - 16/ π²) i + (16/ π² – 24/π) j ft/sec² (result)