Problem 2/3 Page 27, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: v(t) = B - C t + D t^3/2 and s=s0 when t=0 for a particle moving in a straight line.

Find: s, v, and a at t=t1.

Relevant variables:
B = 2 m/s
C = 4 m/s²
D = 5 m/s^5/2
s0=3 m
t1=3 s

Finding v(t1):

v(t1) = B - C t1 + D t1^3/2 = 15.98 m/s

Finding s(t1):
v = ds/dt = B - C t + D t^3/2
Separating variables:

ds = { B - C t + D t^3/2 } dt

Integrating both sides:

s = B t - 1/2 C t² + 2/5 t^5/2 + F

Noting that s is known at t=0, permits the evaluation of F.

s = B t - 1/2 C t² + 2/5 D t^5/2 + s0 = 22.2 m

Evaluating at t1:

s(t1) = B t1 - 1/2 C t1² + 2/5 t1^5/2 + s0 =

Finding a(t1):

a = dv/dt

a = -C + 3/2 D t^1/2

a(t1) = -C + 3/2 D t1^1/2 = 8,99 m/s²